Nm hc 2010-2011 s 1Thi gian lm bi: 120 phtCu1: (2 im)Cho dy t s bng nhau: 2 2 2 2 a b c d a b c d a b c d a b c da b c d+ + + + + + + + + + + + Tm gi tr biu thc: M= a b b c c d d ac d d a a b b c+ + + ++ + ++ + + +Cu2: (1 im) .Cho S=abc bca cab + +.Chng minh rng S khng phi l s chnh phng.Cu3: (2 im)Mt t chy t A n B vi vn tc 65 km/h, cng lc mtxe my chy t B n A vi vn tc 40 km/h. Bit khong cch AB l 540 km v M l trung im ca AB. Hi sau khi khi hnh bao lu th t cch M mt khong bng 1/2 khong cch t xe my n M.Cu4: (2 im)Cho tam gic ABC, O l im nm trong tam gic.a. Chng minh rng: BOC A ABO ACO + +b. Bit 0902AABO ACO + v tia BO l tia phn gic ca gc B. Chng minh rng: Tia CO l tia phn gic ca gc C.Cu 5: (1,5im).Cho 9 ng thng trong khng c 2 ng thng no song song. CMR t nht cng c 2 ng thng m gc nhn gia chng khng nh hn 200.Cu 6: (1,5im).Khi chi c nga, thay v gieo 1 con sc sc, ta gieo c hai con sc sc cng mt lc th im thp nht l 2, cao nht l 12. cc im khc l 3; 4; 5 ;6 11. Hy lp bng tn s v kh nng xut hin mi loi im ni trn? Tnh tn xut ca mi loi im .------------------------------------ Ht ---------------------------------------------- s 2.Thi gian lm bi: 120 phtCu 1: Tm cc s a,b,c bit rng: ab =c ;bc= 4a; ac=9bTrang 1Nm hc 2010-2011Cu 2:Tm s nguyn x tho mn:a, 5x-3< 2 b, 3x+1>4 c,4- x+2x =3Cu3: Tm gi tr nh nht ca biu thc:A = x+ 8 -xCu 4: Bit rng :12+22+33+...+102= 385.Tnh tng : S= 22+ 42+...+202Cu 5 :Cho tam gic ABC ,trung tuyn AM .Gi I l trung im ca on thng AM, BI ct cnh AC ti D.a. Chng minh AC=3 ADb. Chng minh ID =1/4BD------------------------------------------------- Ht ------------------------------------------ s 3Thi gian lm bi: 120 phtCu 1 . ( 2) Cho: dccbba . Chng minh:dad c bc b a ,_

+ ++ +3.Cu 2. (1). Tm A bit rng: A = a c bb a cc b a+++.Cu 3. (2). Tm Z x A Z v tm gi tr .a). A = 23+xx.b). A =32 1+xx.Cu 4.(2).Tm x, bit:a)3 x = 5 . b).( x+ 2) 2 = 81.c). 5 x + 5 x+ 2 = 650Cu 5.(3). Cho ABC vung cn ti A, trung tuyn AM . E BC, BH AE, CK AE, (H,K AE). Chng minh MHK vung cn.-------------------------------- Ht ------------------------------------ s 4Thi gian lm bi : 120 pht.Cu 1 : ( 3 im).1. Ba ng cao ca tam gic ABCc di l 4,12 ,a . Bit rng a l mt s t nhin. Tm a ?2. Chng minh rng t t l thc dcba ( a,b,c ,d0, a b, c d) ta suy ra c cc t l thc:Trang 2Nm hc 2010-2011a) d c cb a a. b) dd cb b a ++.Cu 2: ( 1 im). Tm s nguyn x sao cho: ( x2 1)( x2 4)( x2 7)(x2 10) < 0.Cu 3: (2 im).Tm gi tr nh nht ca:A = x-a+ x-b+ x-c+ x-dvi a 1. c.2 3 x + 5.Cu2: ( 2 im)a. Tnh tng: A= (- 7) + (-7)2 + + (- 7)2006 + (- 7)2007. Chng minh rng: A chia ht cho 43.b. Chng minh rng iu kin cn v m2 + m.n + n2 chia ht cho 9 l: m, n chia ht cho 3.Trang 8Nm hc 2010-2011Cu 3: ( 23,5 im) di cc cnh ca mt tam gic t l vi nhau nh th no,bit nu cng ln lt di tng hai ng cao ca tam gic th cc tng ny t l theo 3:4:5.Cu 4: ( 3 im ) Cho tam gic ABC cn ti A. D l mt im nm trong tam gic, bitADB> ADC . Chng minh rng: DB < DC.Cu 5: ( 1 im ) Tm GTLN ca biu thc: A = 1004 x - 1003 x +.-------------------------------------- Ht --------------------------------- s 14Thi gian : 120Cu 1 (2 im): Tm x, bit : a. 3x 2 +5x = 4x-10 b. 3+ 2x 5 +> 13Cu 2: (3 im )a.Tm mt s c 3 ch s bit rng s chia ht cho 18 v cc ch s ca n t l vi 1, 2, 3.b. Chng minh rng:Tng A=7 +72+73+74+...+74n chia ht cho 400(nN). Cu 3 : (1im )cho hnh v , bit ++ = 1800chng minh Ax// By. A x

C B yCu 4 (3 im ) Cho tam gic cn ABC, c ABC=1000. K phn gic trong ca gc CAB ct AB ti D. Chng minh rng:AD + DC =ABCu 5 (1 im )Tnh tng.S = (-3)0 + (-3)1+ (-3)2 + .....+ (-3)2004.------------------------------------ Ht ---------------------------------- s 15Thi gian lm bi: 120 phTrang 9Nm hc 2010-2011Bi 1: (2,5) Thc hin php tnh sau mt cch hp l:1 1 1 1 1 1 1 1 190 72 56 42 30 20 12 6 2 Bi 2: (2,5)Tnh gi tr nh nht ca biu thc: A = x x + 5 2Bi 3: (4) Cho tam gic ABC. Gi H, G,O ln lt l trc tm , trng tm v giao im ca 3 ngtrung trc trong tam gic. Chng minh rng:a. AH bng 2 ln khong cch t O n BCb. Ba im H,G,O thng hng v GH= 2 GOBi 4: (1 ) Tm tng cc h s ca a thc nhn c sau khi b du ngoc trong biu thc (3-4x+x2)2006.(3+ 4x + x2)2007.------------------------------------------- Ht ------------------------------------------ 16Thi gian lm bi: 120 phtCu 1(3): Chng minh rngA = 22011969 + 11969220 + 69220119 chia ht cho 102Cu 2(3): Tm x, bit: a. x x 2 3 + + ; b.3x 5 x 2 +Cu 3(3): Cho tam gic ABC. Gi M, N, P theo th t l trung im ca BC, CA, AB. Cc ng trung trc ca tam gic gp nhau tai 0. Cc ng cao AD, BE, CF gp nhau ti H. Gi I, K, R theo th t l trung im ca HA, HB, HC.a) C/m H0 v IM ct nhau ti Q l trung im ca mi on.b) C/m QI = QM = QD = 0A/2c) Hy suy ra cc kt qu tng t nh kt qu cu b.Cu 4(1):Tm gi tr ca x biu thcA = 10 - 3|x-5|t gi tr ln nht.--------------------------------------------- Ht --------------------------------------------- 17Thi gian: 120 phtBi 1: (2) Cho biu thcA = 35+xxTrang 10Nm hc 2010-2011a) Tnh gi tr ca A ti x = 41b) Tm gi tr ca x A = - 1c) Tm gi tr nguyn ca x A nhn gi tr nguyn.Bi 2. (3)a) Tm x bit:1 7 x xb) Tnh tng M = 1 + (- 2) + (- 2)2 + +(- 2)2006c) Cho a thc: f(x) = 5x3 + 2x4 x2 + 3x2 x3 x4 + 1 4x3. Chng t rng a thc trn khng c nghimBi 3.(1)Hi tam gic ABC l tam gic gbit rng cc gc ca tam gic t l vi 1, 2, 3.Bi 4.(3) Cho tam gic ABC c gc B bng 600. Hai tia phn gic AM v CN ca tam gic ABC ct nhau ti I.a) Tnh gc AICb) Chng minh IM = INBi 5. (1)Cho biu thcA = x x62006.Tm gi tr nguyn ca x A t gi tr ln nht. Tm gi tr ln nht .---------------------------------------- Ht -------------------------------------- 18Thi gian: 120 phtCu 1:1.Tnh:a. 20 1521

,_

,_

41. b. 30 2591

,_

,_

31:2. Rt gn:A = 20 . 6 3 . 26 . 2 9 . 48 8 109 4 5+3. Biu din s thp phn di dng phn s v ngc li:a. 337b. 227c. 0, (21) d. 0,5(16)Cu 2:Trong mt t lao ng, ba khi 7, 8, 9 chuyn ch c 912 m3 t. Trung bnh mi hc sinh khi 7, 8, 9 theo th tlm -c 1,2 ; 1,4 ; 1,6 m3 t. S hc sinh khi 7, 8 t l vi 1 v 3. Khi 8 v 9 t l vi 4 v 5. Tnh s hc sinh mi khi.Trang 11Nm hc 2010-2011Cu 3:a.Tm gi tr ln nht ca biu thc:A = 4 ) 2 (32+ + xb.Tm gi tr nh nht ca biu thc:B = (x+1)2 + (y + 3)2 + 1Cu 4: Cho tam gic ABC cn (CA = CB) v C = 800. Trong tam gic sao cho 0MBA 30 v010 MAB .TnhMAC.Cu 5:Chng minh rng : nu (a,b) = 1 th (a2,a+b) = 1.------------------------------------- Ht -------------------------------------- 19 Thi gian: 120 pht.Cu I: (2)1) Cho 654 32 1 + c b a v 5a - 3b - 4 c = 46 . Xc nh a, b, c2) Cho t l thc : dcba . Chng minh : cd dd cd cab bb ab a3 25 3 23 25 3 222 222 2++ ++ . Vi iu kin mu thc xc nh.Cu II : Tnh : (2)1) A = 99 . 971....7 . 515 . 31+ + +2) B = 51 50 3 23131.....313131 + + + Cu III : (1,5 )i thnh phn s cc s thp phn sau :a. 0,2(3) ;b.1,12(32).Cu IV : (1.5) Xc nh cc a thc bc 3 bit :P(0) = 10; P(1) = 12; P(2) = 4 ; p(3) = 1Cu V : (3)Cho tam gic ABC c 3 gc nhn. Dng ra pha ngoi 2 tam gic vung cn nh A l ABD v ACE . Gi M;N;P ln lt l trung im ca BC; BD;CE .a. Chng minh : BE = CD v BE vi CDb. Chng minh tam gic MNP vung cn---------------------------------------------- Ht ------------------------------------------------- 20Thi gian lm bi: 120 phtTrang 12Nm hc 2010-2011Bi 1 (1,5): Thc hin php tnh:a) A = 3 30,375 0,31,5 1 0,7511 125 5 50,265 0,5 2,5 1,2511 12 3 + ++ + + + b) B = 1 + 22 + 24 + ... + 2100Bi 2 (1,5):a) So snh: 230 + 330 + 430 v 3.2410b) So snh: 4 +33 v29+ 14Bi 3 (2):Ba my xay xay c 359 tn thc. S ngy lm vic ca cc my t l vi 3:4:5, s gi lm vic ca cc my t l vi 6, 7, 8, cng sut cc my t l nghc vi 5,4,3. Hi mi my xay c bao nhiu tn thc.Bi 4 (1):Tm x, y bit:a) 3 4 x 3 b) 1 1 1 1... 21.2 2.3 99.100 2x _+ + + ,Bi 5 ( 3):ChoABC c cc gc nh hn 1200. V pha ngoi tam gic ABC cc tam gic u ABD, ACE. Gi M l giao im ca DC v BE. Chng minh rng:a) 0120 BMC b) 0120 AMBBi 6 (1):Cho hm s f(x) xc nh vi mi x thuc R. Bit rng vi mi x ta u c:21( ) 3. ( ) f x f xx+ . Tnh f(2).---------------------------------------- Ht ------------------------------------------ 21Thi gian lm bi: 120 phtCu 1 (2) Tm x, y, z Z, bita. x x + = 3 - xb.21 16 yxc. 2x = 3y; 5x = 7z v 3x - 7y + 5z = 30Cu 2 (2)Trang 13Nm hc 2010-2011a. Cho A =) 11001)...( 141).( 131).( 121(2 2 2 2 .Hy so snh A vi21b. Cho B = 31+xx .Tm x Z B c gi tr l mt s nguyn dngCu 3 (2)Mt ngi i t A n B vi vn tc 4km/h v d nh n B lc 11 gi 45 pht. Sau khi i c 51 qung ng th ngi i vi vn tc 3km/h nn n B lc 12 gi tra.Tnh qung ngAB v ngi khi hnh lc my gi?Cu 4 (3) ChoABC c A > 900. Gi I l trung im ca cnh AC. Trn tia i ca tia IB ly im D sao cho IB = ID. Ni c vi D.a. Chng minhCID AIB b. Gi M l trung im ca BC; N l trung im ca CD. Chng minh rng I l trung im ca MNc. Chng minh AIB AIB BIC a2 + b2 + c2.Bi 5:(3 im) Cho tam gic ABC c 0B=C=50. Gi K l im trong tam gic sao cho 0 0KBC=10 KCB=30a. Chng minh BA = BK.b. Tnh s o gc BAK.--------------------------------- Ht ----------------------------------Trang 16Nm hc 2010-2011 thi 26Thi gian lm bi: 120 phtCu 1.Vi mi s t nhin n 2hy so snh:a. A= 2 2 2 21....413121n+ + + + vi 1 .b. B = ( )2 2 2 221...614121n+ + + + vi 1/2Cu 2:Tm phn nguyn ca, vi 1 4 31....34232+++ + + + nnnCu 3:Tm t l 3 cnh ca mt tam gic, bit rng cng ln lt di hai ng cao ca tam gic th t l cc kt qu l5: 7 : 8.Cu 4:Cho gc xoy , trn hai cnh ox v oy ln lt ly cc im A v B cho AB c di nh nht.Cu 5:Chng minh rng nu a, b, c vc b a + +l cc s hu t.--------------------------------------------------------------Phn 2: H ng dn giiHng dn gii s 1.Cu 1:Mi t s cho u bt i 1 ta c:2 21 1a b c d a b c da b+ + + + + + =2 21 1a b c d a b c dc d+ + + + + + a b c d a b c d a b c d a b c da b c d+ + + + + + + + + + + + +, Nu a+b+c+d 0tha = b = c = d lc M = 1+1+1+1=4+, Nu a+b+c+d = 0tha+b = - (c+d); b+c = - (d+a); c+d = - (a+b);Trang 17AMBNm hc 2010-2011d+a = -(b+c), lc M = (-1) + (-1)+ (-1) + (-1) =-4.Cu 2: S = (100a+10b+c)+(100b+10c+a)+ (100c+10a+b) = 111(a+b+c) = 37.3(a+b+c).V 0 < a+b+c27 nn a+b+c/ M37.Mt khc( 3; 37) =1 nn 3(a+b+c) M37 => S khng th l s chnh phng.Cu 3:Qung ng AB di 540 Km; na qung dng AB di 270 Km. Gi qung ng t v xe my i l S1, S2. Trong cng 1 thi gian th qung ng t l thun vi vn tc do 1 21 2S StV V (t chnh l thi gian cn tm).t= 270 270 2 540 2 270 2 (540 2 ) (270 2 ) 270; 365 40 130 40 130 40 90a a a a a at Vy sau khi khi hnh 3 gi th t cch M mt khong bng 1/2 khong cch t xe my n M.Cu 4:a,Tia CO ct AB ti D.+, Xt BOD c BOC l gc ngoi nn BOC = 1 1B D ++, Xt ADC c gc D1 l gc ngoi nn 1 1D A C +Vy BOC = 1A C + +1Bb, Nu 0902AABO ACO + th BOC = 0 090 902 2A AA+ +Xt BOC c: ( ) 0 0 02 200 02180 180 902 218090 902 2 2A BC O BA B C CC _ + + + ,+ tia CO l tia phn gic ca gc C.Cu 5:Ly im O tu .Qua O v 9 ng thng ln lt song song vi 9 -ng thng cho. 9 ng thng qua O to thnh 18 gc khng c im trong chung, mi gc ny tng ng bng gc gia hai ng thng trong s 9 ng thng cho. Tng s o ca 18 gc nh Trang 18ABCDONm hc 2010-2011O l 3600 do t nht c 1 gc khng nh hn 3600 : 18 = 200, t suy ra t nht cng c hai ng thng m gc nhn gia chng khng nh hn 200.Cu 6:Tng s im ghi hai mt trn ca hai con sc sc c th l:2 = 1+13 = 1+2 = 2+14 = 1+3 =2 +2 = 3+15 = 1+4 =2+3=3+2=4+1.6=1+5=2+4=3+3=4+2=5+17=1+6=2+5=3+4= 4+3=5+2=-6+18= 2+6=3+5=4+4=5+3=6+29=3+6=4+5=5+4=6+310=4+6=5+5=6+411=5+6=6+512=6+6. im s (x) 2 3 4 5 6 7 8 9 10 11 12Tn s( n) 1 2 3 4 5 6 5 4 3 2 1Tn sut (f) 2,8% 5,6% 8,3% 11,1% 13,9% 16,7% 13,9% 11,1% 8,3% 5,6% 2,8%Nh vy tng s 7 im c kh nng xy ra nht ti 16,7%-------------------------------------------------------------------p n s 2Cu1: Nhn tng v bt ng thc ta c : (abc)2=36abc+, Nu mt trong cc s a,b,c bng 0 th 2 s cn li cng bng 0+,Nuc 3s a,b,c khc 0 th chia 2 v cho abc ta c abc=36+, T abc =36 v ab=c ta c c2=36 nn c=6;c=-6+, T abc =36 v bc=4a ta c 4a2=36 nn a=3; a=-3+, T abc =36 v ab=9b ta c 9b2=36 nn b=2; b=-2-, Nu c = 6 th av b cng du nn a=3, b=2 hoc a=-3 , b=-2-, Nu c = -6 th av b tri du nn a=3 b=-2 hoc a=-3 b=2Trang 19Nm hc 2010-2011Tm li c 5 b s (a,b,c) tho mn bi ton(0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)Cu 2. (3)a.(1) ( 5x-3( -2 x>1*Nu 3x+1 x1 hoc x x 4 (0,25)(1)4-x+2x=3 => x=-1( tho mn k) (0,25)*4-x x>4 (0,25)(1) x-4+2x=3 x=7/3 (loi) (0,25)Cu3. (1) p dng ( a+b(( a( +( b( Ta cA=( x( +( 8-x( ( x+8-x( =8MinA =8 x(8-x) 0 (0,25)*' 0 80xx=>0 x 8 (0,25)*' 0 80xx=> '80xxkhng tho mn(0,25)Vy minA=8 khi 0 x 8(0,25)Cu4.Ta c S=(2.1)2+(2.2)2+...+ (2.10)2(0,5) =22.12+22.22+...+22.102=22(12+22+...+102) =22.385=1540(0,5)Cu5.(3)Chng minh: a (1,5)Trang 20AB MCDENm hc 2010-2011Gi E l trung im CD trong tam gic BCD c ME l ng trung bnh => ME//BD(0,25)Trong tam gic MAE c I l trung im ca cnh AM (gt) m ID//ME(gt)Nn D l trung im ca AE => AD=DE (1)(0,5)V E l trung im ca DC => DE=EC (2) (0,5)So snh (1)v (2) => AD=DE=EC=> AC= 3AD(0,25)b.(1)Trong tam gic MAE ,ID l ng trung bnh (theo a) => ID=1/2ME (1) (0,25)Trong tam gic BCD; ME l ng trung bnh => ME=1/2BD (2)(0,5)So snh (1) v (2) => ID =1/4 BD (0,25)----------------------------------------------------------------p n s 3Cu 1.Ta c. . .dadccbba (1) Ta li c.a c bc b adccbba+ ++ + (2)T (1) v(2) => dad c bc b a ,_

+ ++ +3.Cu 2. A = a c bb a cc b a+++.= ( ) c b ac b a+ ++ +2.Nu a+b+c 0 => A = 21.Nua+b+c = 0=>A = -1.Cu 3.a). A = 1 + 25 x A Z thx- 2 l c ca 5.=> x 2 = (t1; t 5)*x = 3=>A = 6*x = 7=>A = 2*x = 1=>A = - 4*x = -3=>A = 0 b) A = 37+ x - 2 A Z thx+ 3 l c ca 7.=> x + 3 = (t1; t 7)*x = -2=>A = 5*x = 4=>A = -1*x = -4 =>A = - 9*x = -10=>A = -3 .Cu 4.a). x =8hoc - 2 b). x =7hoc - 11c). x =2.Trang 21Nm hc 2010-2011Cu 5. ( T v hnh) MHK l cnti M .Tht vy: ACK = BAH. (gcg)=>AK = BH . AMK = BMH (g.c.g) =>MK = MH.Vy: MHK cnti M .--------------------------------------------------------------------p n s 4Cu 1:Gi x, y, z l di 3 cnh tng ngvi cc ng cao bng 4, 12, a. Ta c: 4x= 12y = az= 2Sx= S/2 ; y = S/6;z = 2S/a(0,5 im)Do x-y < z< x+y nn32 2626 226 2< < + < < aS SaS S S(0,5 im) 3, a , 6 Do a Nnn a=4 hoc a= 5. (0,5 im)2. a. Tdcba d c cb aad cb acad cb adbca (0,75 im)b. dcbadd cb b ad cb adbd cb adbca ++++ ++ (0,75 im)Cu 2: V tch ca 4 s :x2 1 ; x2 4; x2 7;x2 10 l s m nn phi c1 s m hoc 3 s m.Ta c :x2 10< x2 7< x2 4< x2 1. Xt 2 trng hp:+ C 1 s m:x2 10 1 2 32 3 4x y z v 2x + 3y - z = 50(0,5)=> x = 11, y = 17, z = 23. (0,5)Cu 3(2): Cc phn s phi tm l: a, b, c ta c : a + b + c = 21370v a : b : c = 345: : 6: 40: 255 1 2 (1) => 9 12 15, ,35 7 14a b c (1)Cu 4(3):K DF // AC ( F thuc BC ) (0,5 )=> DF = BD = CE (0,5 ) =>IDF = IFC ( c.g.c ) (1 )=> gc DIF = gc EIC => F, I, C thng hng => B, I, C thng hng(1)Cu 5(1):=> 7.2 1 1(14 1) 77xy xy+ + => (x ; y ) cn tm l ( 0 ; 7 )----------------------------------------------------------------------p n s 6:Cu 1: a)Ta c: 21112 . 11 ; 31213 . 21 ;41314 . 31 ; ;1001991100 . 99 1 Vy A = 1+10099100111001991991....31312121 ,_

++ + ,_

++ ,_

+b) A = 1+

,_

+ + ,_

+ ,_

+ ,_

221 . 20201....25 . 44124 . 33123 . 221== 1+ ( ) + + + + + + + 21 ... 4 3 221221...2423=

,_

1222 . 2121= 115.Cu 2: a) Ta c:4 17 > ; 5 26 >nn 1 5 4 1 26 17 + + > + + hay 10 1 26 17 > + +Cn99 < 10 .Do : 99 1 26 17 > + +b) ;10111>10121>;10131>; ..; 1011001.Trang 24Nm hc 2010-2011Vy: 10101. 1001001....312111 > + + + +Cu 3: Gi a,b,ca l cc ch s ca s c ba ch scn tm . V mi ch s a,b,ca khng vt qu 9 v ba ch s a,b,ca khng th ng thi bng 0 , v khi ta khng c s c ba ch s nn: 1 a+b+c27Mt khc s phi tm l bi ca 18 nn a+b+c =9 hoc a+b+c = 18 hoc a+b+c=17Theo gi thit, ta c:6 3 2 1c b a c b a + + Do : ( a+b+c) chia ht cho 6Nn : a+b+c =18 36183 2 1 c b a a=3; b=6 ; ca =9V s phi tm chia ht cho 18 nnch s hng n v ca n phi l s chn.Vy cc s phi tm l: 396; 936.Cu 4:a) V AH BC; ( H BC) ca ABC+ hai tam gic vung AHB v BIDc:BD= AB (gt)Gc A1= gc B1( cng ph vi gcB2) AHB= BID ( cnh huyn, gc nhn)AH BI (1) v DI= BH+ Xt hai tam gic vung AHC v CKE c: Gc A2= gc C1( cng ph vi gc C2)AC=CE(gt) AHC= CKB ( cnh huyn, gc nhn) AH= CK (2)t (1) v (2) BI= CK v EK = HC.b) Ta c: DI=BH ( Chng minh trn)tng t: EK = HCT BC= BH +Hc= DI + EK.Cu 5: Ta c:A = 1 2001 + x x=2000 1 2001 1 2001 + + x x x xTrang 25Nm hc 2010-2011Vy biu thc cho t gi tr nh nht l 2000 khi x-2001 v 1-x cng du, tc l :1x 2001biu im :Cu 1: 2 im . a. 1 imb. 1imCu 2: 2 im : a. 1 im b . 1 im .Cu 3 : 1,5 imCu 4: 3 im :a. 2 im ; b. 1 im .Cu 5 : 1,5 im .---------------------------------------------------------------------p n s 7Cu1:a,(1)0 4534913245132541326313272 ++ +++ +++ +++ ++x x x x x (0,5 )...... 0 )513241325132613271)( 329 ( + + + + + x329 0 329 + x x (0,5 )b, a.Tm x,bit: |5x - 3| - x = 7 5 3 7 x x + (1)(0,25 )K:x -7(0,25 )( )( )5 3 715 3 7x xx x + +.(0,25 )Vy c hai gi tr x tha mn iu kin u bi. x1 = 5/2;x2= - 2/3 (0,25).Cu 2:a, 2007 4 3 271.....717171711 + + + S ; 2006 3 271.....7171711 7 7 + + S(0.5) 2007717 8 S

87172007 S(0,5)b,! 100 1 100.......! 3 1 3! 2 1 2! 10099......! 43! 32! 21 + ++ + + + +(0,5)................... 1! 10011 < (0,5)c, Ta c +23n) 2 2 ( 3 3 2 3 22 2 2 n n n n n n n + ++ + +(0,5)................. ( ) 10 2 3 10 10 . 2 10 . 3 5 . 2 10 . 32 2 n n n n n n(0,5)Trang 26Nm hc 2010-2011Cu 3: Gi di 3 cnh l a , b, c, 3 chiu cao tng ng l x, y, z, din tch S ( 0,5 )xSa2

ySb2 zSc2 (0,5) zSySxS c b a4232224 3 2 (0,5)3 4 64 3 2z y xz y x vy x, y, z t l vi6 ; 4 ; 3(0,5)Cu4: GT; KL; Hnh v(0,5)a,Gc AIC = 1200 (1 )b,Ly AC H : AH = AQ .............. IP IH IQ (1 )Cu5:B ; LN ( ) 3 1 2 ;2+ n LN BNNV( ) ( ) 3 3 1 2 0 12 2 + n n t NN khi bng 3 (0,5)Du bng xy ra khi1 0 1 n nvy B ; LN 31 B v1 n (0,5)-------------------------------------------------------------p n s 8Cu 1 : 3 im . Mi cu 1 ima) (x-1)5 = (-3)5 x-1 = -3 x = -3+1 x = -2b) (x+2)(151141131121111 + +) = 0151141131121111 + + 0 x+2 = 0 x = 2c) x - 2 x= 0 ( x )2- 2 x= 0 x ( x - 2) = 0 x= 0 x = 0hocx - 2 = 0 x= 2 x = 4Cu 2 : 3 im . Mi cu 1,5 ima) 8145 + yx,8182 5 +yx ,82 1 5 yxx(1 - 2y) = 40 1-2y l c l ca 40 . c l ca 40 l :t1 ;t5 .p s :x = 40 ; y = 0x = -40 ; y = 1x = 8 ; y = -2x = -8 ; y = 3b) Tm xz AZ. A= 34131+ +x xxTrang 27Nm hc 2010-2011A nguyn khi 34 x nguyn 3 x(4) = {-4 ; -2 ;-1; 1; 2; 4}Cc gi tr ca x l : 1 ; 4; 16 ; 25 ; 49 .Cu 3 : 1 im23 5 x - 2x = 143 5 x=x + 7 (1)K:x -7(0,25 )( )( )5 3 715 3 7x xx x + +.(0,25 )Vy c hai gi tr x tha mn iu kin u bi. x1 = 5/2;x2= - 2/3 (0,25). Cu4.(1.5 im)Cc gc A, B , C t l vi 7, 5, 3121518015 3 5 70 + + C B A C B AA= 840

gc ngoi ti nh A l 960B =600 gc ngoi ti nh B l 1200C = 360 gc ngoi ti nh C l 1440 Cc gc ngoi tng ng t l vi 4 ; 5 ; 6b)1) AE = AD ADE cn 1E DE EDA 1E = 01802A (1) ABC cn B C 1AB C= 01802A (2)T (1) v (2) 1E ABC ED // BCa) XtEBC vDCB c BC chung (3) EBC DCB (4)BE = CD (5)T (3), (4), (5) EBC =DCB (c.g.c) BEC CDB = 900 CE AB ..p n s 9Trang 28Nm hc 2010-2011Bi 1: 3 ima, Tnh:A = 11160.36471300475.11121 .3311 1160).41915(100175310(1112)71767183(331 = 1815284284551001.332841001553357 34110011001100110561119331 b, 1,5 im Ta c:+) 1 + 4 +7 ++ 100= ( 1+100) + ( 4 + 97) +.+ ( 49+ 52)= 101 . 34 = 143434 cp+) 1434 410 = 1024+) ( 18 . 123 + 9 . 436 . 2 + 3 . 5310. 6 )= 18 . ( 123 + 436 + 5310 )= 18 . 5869 =105642Vy A = 105642 : 1024103,17Bi 2: 2 imGii s cn tm l x, y, z. S nh l x , s ln nht l z. Ta c: xy z (1)Theo gi thit:21 1 1 + +z y x(2). Do (1) nn z =x z y x3 1 1 1 + +Vy: x = 1. Thay vo (2) , c: y z y211 1 +Vy y = 2. T z = 2.Ba s cn tm l 1; 2; 2.Bi 3:2 imC 9 trang c 1 ch s. S trang c 2 ch s l t 10 n 99 nn c tt c 90 trang. Trang c 3 ch s ca cun sch l t 100 n 234, c tt c 135 trang. Suy ra s cc ch s trong tt c cc trang l:9 + 2 . 90 + 3. 135 = 9 + 180 + 405 = 594Bi 4 : 3 imTrn tia EC ly im D sao cho ED = EA.Hai tam gic vungABE =DBE ( EA = ED, BE chung)Suy ra BD = BA ; BAD BDA .Theo gi thit: EC EA = A BVy EC ED = AB Hay CD = AB (2)T (1) v (2) Suy ra: DC = BD.Trang 29Nm hc 2010-2011V tia ID l phn gic ca gc CBD ( I BC ).Hai tam gic:CID vBID c :ID l cnhchung,CD = BD ( Chng minh trn). CID=IDB ( v DI l phn gic ca gc CDB )VyCID = BID ( c . g . c) C =IBD. Gi C l BDA=C+IBD= 2

C = 2 ( gc ngoi ca BCD) m A=D( Chng minh trn) nn A= 2 + 2 = 900

= 300 .Do ; C=300 v A = 600----------------------------------------------H ng dn gii s 9 Bi 1.a.Xt 2 trng hp :*5 x ta c : A=7.* 5 x > hay A > 7.Vy : Amin = 7 khi5 x .Bi 2. a.t : A =2 2 2 21 1 1 1.......5 6 7 100+ + + +Ta c :*A < 1 1 1 1.........4.5 5.6 6.7 99.100+ + + + = 1 1 1 1 1 1.....4 5 5 6 99 100 + + + = 1 1 14 100 4 1 1 1 1 1 1 1.........5.6 6.7 99.100 100.101 5 101 6+ + + + >.b.Ta c : 2 9 5 17 33 3 3a a aa a a+ ++ + + += 4 263aa++ == 4 12 14 4( 3) 14 1443 3 3a aa a a+ + + + ++ + + l s nguynKhi (a + 3) l c ca 14 m (14) = 1;2;7;14 t t t t.Ta c : a = -2;- 4;- 1; - 5; 4 ; - 10; 11 ; -17.Bi 3. Bin i :( ) 12 1 30. A n nn + + ( ) 6 1 30 6 A n nn n1 + ]M M* ( ) 1 30 nn n n M M n (30) hay n {1, 2 , 3, 5 , 6 , 10 , 15 , 30}.* ( ) ( ) 30 6 16 1 3 nn nn M M M+ { } 3 3, 6,15, 30 . n n MTrang 30Nm hc 2010-2011+( ) { } 1 3 1,10 . n n M n {1 , 3 , 6 , 10 , 15 , 30}.-Th tng trng hp ta c : n = 1, 3, 10, 30 tho mn bi ton.Bi 4.-Trn Oy ly M sao cho OM = m. Ta c :N nm gia O, M v MN = OM.-Dng d l trung trc ca OM v Oz lphn gic ca gc xOy chng ct nhau ti D.-' ( . . ) ODM MDNc g c MD ND V VD thuc trung trc ca MN.-R rng : D c nh. Vy ng trung trc ca MN i qua D c nh.Bi 5. -Dng tng qut ca a thc bc hai l :( )2f x ax bx c + +(a0).- Ta c :( ) ( ) ( )21 1 1 f x a x b x c + + .- ( ) ( ) 1 2 f x f x ax a b x + 2 10ab a ' 1212ab 'Vy a thc cn tm l :( )21 12 2f x x x c + + (c l hng s).p dng :+ Vi x = 1 ta c :( ) ( ) 1 1 0 . f f + Vi x = 2 ta c :( ) ( ) 1 2 1 . f f .+ Vi x = n ta c :( ) ( ) 1 . n f n f n S = 1+2+3++n =( ) ( ) 0 f n f = ( )212 2 2nnn nc c++ + .L u :Hc sinh gii cch khc ng vn cho im ti a. Bi hnh khng v hnh khng chm im.--------------------------------------------------------------------p n s 11Cu1 (lm ng c 2 im)Ta c: 228 20x xx x+ =222 10 20x xx x x + = 2( 2)( 10)x xx x + (0,25)Trang 31 xz ddmn i ym'oNm hc 2010-2011iu kin(x-2)(x+10) 0 x 2;x -10 (0,5)Mt khc 2 x = x-2nu x>2-x + 2 nux< 2(0,25)* Nu x> 2 th 2( 2)( 10)x xx x +=( 2)( 2)( 10)xxx x + = 10xx +(0,5)* Nu x 0; y >0 ; z >0)Theo ra ta c{94(1)3 4 5 (2)x y zx y z++ (0,5)BCNN (3,4,5) = 60T (2) 360x=460y=560zhay 20x=15y=12z(0,5)p dng tnh cht dy t s bng nhau ta c :20x=15y=12z= 20 15 12x y z + ++ += 9447=2(0,5)x= 40, y=30 v z =24 (0,5)S hc sinh i trng cy ca 3 lp 7A, 7B, 7C ln lt l 40, 30, 24.Cu 3 (lm ng cho 1,5) 200610 539+ l s t nhin 102006 + 53 M9(0,5) 102006 + 53 M9102006 + 53c tng cc ch s chia ht cho 9m 102006 + 53= 1+ 0 +0 +.........+0 + 5+3= 9M9 102006 + 53 M9hay200610 539+ l s t nhin(1)Cu 4 (3)- V c hnh, ghi GT, KL c 0,25a, ABC c 1 2A A (Az l tia phn gic caA ) 1 1A C (Ay // BC, so le trong)Trang 32Nm hc 2010-2011 2 1A C ABC V cn ti Bm BK AC BK l ng cao ca cn ABC BK cng l trung tuyn ca cn ABC (0,75)hay K l trung im ca ACb, Xt ca cn ABH vvung BAK.C AB l cng huyn(cnh chung) 02 1( 30 ) A B V {020 0 0130290 60 30AAB vung ABH = vung BAK BH = AK m AK = 2 2AC ACBH (1)c, AMC vung ti M c AK = KC = AC/2(1) MK l trung tuyn thuc cnh huyn KM = AC/2(2)T (10 v (2) KM = KC KMC cn.Mt khc AMC c 0 0 0 0 090 A=30 90 30 60 M MKC AMC u (1)Cu 5. Lm ng cu 5 c 1,5Xy dng s cy v gii bi tonp n : Ty t gii nht, Nam gii nh, ng gii 3, Bc gii 4-------------------------------------p n s 12Cu 1: (2)a) Xt khong 32 xc x = 4,5 ph hp 0,25 Xt khong 32< xc x = - 45ph hp 0,25 b) Xt khong 23 x c x > 4 0,2Xt khong 23< x c x < -1 0,2Trang 33Nm hc 2010-2011Vy x > 4 hoc x < -1 0,1c) Xt khong 31 x Ta c 3x - 1 738 x Ta c 3831 xXt khong 31< x Ta c -3x + 172 xTa c 312 xVy gi tr ca x tho mn bi l 382 xCu 2:a) S = 1+25 + 252 +...+ 25100 0,31 25 25 2425 ... 25 25 25101101 2 + + + S S SS 0,3Vy S = 241 25101 0,1b) 430= 230.230 = (23)10.(22)15 >810.315> (810.310)3 = 2410.3 0,8Vy 230+330+430> 3.224 0,2Cu 3:a) Hnh a.AB//EF v c hai gc trong cng pha b nhauEF//CD v c hai gc trong cng pha b nhauVy AB//CDb) Hnh b.AB//EF V c cp gc so le trong bng nhau 0,4CD//EF v c cp gc trong cng pha b nhau 0,4Vy AB//CD 0,2Cu 4:(3)Trang 34Nm hc 2010-2011a)MN//BC MD//BD D trung im AP 0,3 BP va l phn gic va l trung tuyn nn cng l ng cao BDAP 0,2Tng t ta chng minh c BE AQ 0,5 b) AD = DPBDE DBP (g.c.g) DP = BE BE = AD 0,5 MD ME c g c MAD MBE ) . . ( 0,3BP = 2MD = 2ME = BQVy B l trung im ca PQ 0,2c)BDE vung B, BM l trung tuyn nn BM = ME 0,4ADB vung D c DM l trung tuyn nn DM = MA 0,4DE = DM + ME = MA + MB 0,2Cu 5: 1A = x +4101A ln nht x 410ln nht0,3Xt x > 4 thx 410 < 0Xt 4 < x th x 410> 0a ln nht 4 - x nh nht x = 3 0,6------------------------------------------------------------------------------p n s 12Cu 1: ( mi 0,5 im ).a/.4 3 x +- x = 15. b/. 3 2 x - x> 1.

4 3 x + = x + 15 3 2 x > x + 1Trang 35Nm hc 2010-2011* Trng hp 1: x -34 , ta c: * Trng hp 1: x 23, ta c:4x + 3 = x + 153x - 2 > x + 1 x = 4 ( TMK). x > 32 ( TMK).* Trng hp 2: x < - 34 , ta c: * Trng hp 2: x < 23, ta c:4x + 3 = - ( x + 15) 3x 2 < - ( x + 1) x = - 185 ( TMK). x < 14 ( TMK)Vy: x = 4 hoc x = - 185. Vy: x > 32 hoc x < 14.c/. 2 3 x + 5 5 2 3 5 x + 4 1 x Cu 2:a/.Ta c: A= (- 7) + (-7)2 + + (- 7)2006 + (- 7)2007 ( 1 ) (- 7)A = (-7)2 + (- 7)3 + + (- 7)2007 + (- 7)2008 ( 2)8A =(- 7) (-7)2008Suy ra:A = 18.[(- 7) (-7)2008 ] = - 18( 72008+ 7 )* Chng minh: A M43.Ta c: A= (- 7) + (-7)2 + + (- 7)2006 + (- 7)2007 , c 2007 s hng. Nhm 3 s lin tip thnh mt nhm (c 669 nhm), ta c:A=[(- 7) + (-7)2 + (- 7)3] + + [(- 7)2005 + (- 7)2006 + (- 7)2007]= (- 7)[1 + (- 7) + (- 7)2] + + (- 7)2005. [1 + (- 7) + (- 7)2]= (- 7). 43 + + (- 7)2005. 43= 43.[(- 7) + + (- 7)2005] M 43Vy : A M 43b/. * iu kin :Nu m M 3 v n M 3 th m2 M 3, mn M 3 v n2 M 3, do : m2+ mn + n2 M 9.* iu kin cn:Ta c: m2+ mn + n2 = ( m - n)2 + 3mn.(*)Trang 36ABCDNm hc 2010-2011Nu m2+ mn + n2 M 9 thm2+ mn + n2 M 3, khi t (*),suy ra: ( m - n)2 M 3 ,do ( m - n) M 3v th( m - n)2 M 9 v 3mn M 9 nn mn M 3 ,do mt trong hai s m hoc n chia ht cho 3 m ( m - n) M 3nn c 2 s m,n u chia ht cho 3.Cu 3:Gi di cc cnh tam gic l a, b, c ; cc ng cao tng ng vi cc cnh l ha , hb , hc .Ta c: (ha +hb) : ( hb + hc ) : ( ha + hc ) = 3 : 4 : 5Hay: 13(ha +hb) = 14( hb + hc ) =15( ha + hc ) = k ,( vi k 0).Suy ra: (ha +hb) = 3k ; ( hb + hc ) = 4k ; ( ha + hc ) = 5k .Cng cc biu thc trn, ta c: ha + hb + hc = 6k.T ta c:ha = 2k;hb =k;hc = 3k.Mt khc, gi S l din tchABC V, ta c:a.ha = b.hb =c.hc a.2k = b.k = c.3k3a = 6b = 2cCu 4:Gi s DC khng ln hn DB hay DC DB.* Nu DC = DB thBDC Vcn ti D nn DBC = BCD.Suy ra:ABD = ACD.Khi ta c: ADB V=ADC V(c_g_c) . Do : ADB = ADC ( tri vi gi thit).* Nu DC < DB th trongBDC V , ta c DBC < BCD m ABC = ACB suy ra:ABD >ACD ( 1 ) .XtADB VvACD Vc: AB = AC ; AD chung ; DC < DB.Suy ra: DAC < DAB ( 2 ).Trang 37Nm hc 2010-2011T (1) v (2) trongADB VvACD Vta li c ADB < ADC , iu ny tri vi gi thit.Vy:DC > DB.Cu 5: ( 1 im)p dng bt ng thc: x y x- y, ta c:A = 1004 x -1003 x + ( 1004) ( 1003) x x + = 2007Vy GTLN ca A l: 2007.Du = xy ra khi: x -1003.-----------------------------------------------------------------H ng dn chm 13 Cu 1-a (1 im ) Xt 2 trng hp 3x-2 0. 3x -2 kt lun : Khng c gi tr no ca x tho mn.b-(1 im ) Xt 2 trng hp 2x +5 0 v 2x+5 kt lun.Cu 2-a(2 im ) Gi s cn tm l abcabc18=> abc9. Vy (a+b+c)9(1)Ta c : 1 a+b+c27(2)T (1) v (2) suy ra a+b+c =9 hoc 18 hoc 27 (3)Theo bi ra 1a= 2b=3c = 6c b a + + (4)T (3) v (4) => a+b+c=18.v t (4) => a, b, c mabc2 => s cn tm : 396, 936.b-(1 im )A=(7 +72+73+74) + (75+76+77+78) + ...+ (74n-3+ 74n-2+74n-1+74n).= (7 +72+73+74) . (1+74+78+...+74n-4).Trong : 7 +72+73+74=7.400 chia ht cho 400 . Nn A400Cu 3-a (1 im ) T C k Cz//By c : 2C +CBy=2v (gc trong cng pha) (1) 1C +CAx=2v V theo gi thit C1+C2 + + = 4v =3600.Vy Cz//Ax.(2)T (1) v (2) => Ax//By.Cu 4-(3 im)ABC cn, ACB =1000=> CAB = CBA =400.Trn AB ly AE =AD. Cn chng minh AE+DC=AB (hoc EB=DC)Trang 38Nm hc 2010-2011AED cn, DAE = 400: 2 =200.=> ADE =AED = 800 =400+EDB (gc ngoi caEDB)=> EDB =400 => EB=ED (1)Trn AB ly C sao cho AC = AC. C CAD = CAD ( c.g.c)D ACD = 1000 vDCE = 800.VyDCE cn => DC =ED (2)T (1) v (2) c EB=DC.A CEBM DC =DC. Vy AD +DC =AB.Cu 5 (1 im).S=(-3)0+(-3)1 + (-3)2+(-3)3+...+ (-3)2004.-3S= (-3).[(-3)0+(-3)1+(-3)2 + ....+(-3)2004]= (-3)1+ (-3)2+ ....+(-3)2005]-3S-S=[(-3)1 + (-3)2+...+(-3)2005]-(3)0-(-3)1-...-(-3)2005.-4S = (-3)2005 -1. S = 41 ) 3 (2005 =41 32005+---------------------------------------------------------p n 13Bi 1: Ta c : - 2161121201301421561721901 = - (10 . 919 . 818 . 717 . 616 . 515 .. 414 . 313 .. 212 . 11+ + + + + + + +) 1= - (101919181.....413131212111 + + + + + ) 1=- (10111)= 109 0,5Bi 2: A = x x + 5 2Vi x3 0,5Vi 2x5 th A = x-2 x+5 = 3 0,5Vi x>5 th A = x-2 +x 5 = 2x 7 >30,5So snh cc gi tr ca A trong cckhong ta thy gi tr nh nht ca A = 3 2x5 1Bi 3: a. Trn tia i ca tia OC ly im N saoTrang 39AC BOGHNm hc 2010-2011cho ON = OC .Gi M l trung im ca BC.nn OMl ng trung bnh catam gic BNC.Do OM //BN, OM= 21 BNDo OM vung gc BC => NB vung gc BCM AH vung gcvi BC v th NB // AH (1)Tng t AN//BHDo NB = AH. Suy ra AH = 2OM (1)b. Gi I, K theo th t l trung im ca AG v HG thIK l ng trung bnh ca tam gic AGH nn IK// AHIK = 21 AH => IK // OMv IK = OM ; KIG =OMG (so le trong)IGK= MGO nn GK = OG v IGK =MGOBa im H,G, O thng hng1Do GK = OG m GK= 21 HG nn HG = 2GOng thng qua 3 im H, G, O c gi l ng thng le.1Bi 4: Tng cc h s ca mt a thc P(x) bt k bng gi tr ca a thc ti x=1. Vy tng cc h s ca a thc:0,5P(x) = (3-4x+x2)2006 . (3+4x + x2)2007Bng P(1) = (3-4+1)2006 (3+4+1)2007= 00,5------------------------------------------------------------p n 14Cu 1: Ta c:220 0 (mod2) nn 22011969 0 (mod2)119 1(mod2) nn 11969220 1(mod2)Trang 40Nm hc 2010-201169 -1 (mod2) nn 69220119 -1 (mod2)Vy A 0(mod2) hay A M2 (1)Tng t:AM 3 (1)A M17(1)V 2, 3, 17 l cc s nguyn t A M 2.3.17 = 102Cu 2: Tm xa) (1,5)Vi x < -2 x = -5/2 (0,5)Vi -2 x 0 khng c gi tr x no tho mn(0,5)Vi x > 0x = (0,5)b) (1,5) Vi x < -2 Khng c gi tr x no tho mn(0,5)Vi -2 x 5/3 Khng c gi tr x no tho mn(0,5)Vi x > 5/3 x = 3,5(0,5)Bi 3:a) D dng chng minh c IH = 0MAIH // 0M do 0MN = HIK (g.c.g) IEDo : IHQ = M0Q (g.c.g) QH = Q0FH NQI = QM Pb) DIM vung c DQ l ng trungK QOtuyn ng vi cnh huyn nnRQD = QI = QMB D MCNhng QI l ng trung bnh ca 0HA nn c) Tng t: QK = QN = QE = OB/2QR = QP = QF = OC/2Bi 4(1):V 3|x-5| 0 x RDo A = 10 - 3|x-5| 10Vy A c gi tr ln nht l 10 |x-5| = 0 x = 5----------------------------------------------------------------p n 15.Bi 1.iu kin x 0(0,25)a) A = - 79(0,5)Trang 41Nm hc 2010-2011b) 3 + x > 0 A = -1 3 5 x x x = 1 (0,5)c) Ta c: A = 1 - 38+ x.(0,25) A Z th 3 + x l c ca 8 x = {1; 25} khi A = {- 1; 0} (0,5)Bi 2.a) Ta c: 1 7 x x 32 ; 31) 1 ( 70 12 ' ' xx xxx xx(1)b) Ta c: 2M = 2 22 + 23 24 + - 22006 + 22007 (0,25) 3M = 1 + 22007(0,25) M = 31 22007+(0,5)c) Ta c: A = x4 + 2x2 +1 1 vi mi x PCM.(1)Bi 3. Ta c: 00 180301 2 3 6A B C 0 0 0 30 ; 60 ; 90 A B C (0,5)Vy tam gic ABC l tam gic vung ti C (0,5)Bi 4.GT, KL (0,5)a) Gc AIC = 1200 (1)b) Ly H AC sao cho AH = AN (0,5)T chng minh IH = IN = IM(1)Bi 5.A = 1 + x 62000 (0,5)AMax6 x > 0 v nh nht 6 x = 1 x = 5.Vy x = 5 tho mn iu kin bi ton khi A Max= 2001 (0,5)--------------------------------------------------------------------p n 15Trang 42Nm hc 2010-2011Cu 1: (2.5)a. a1.55 40 15 20 152121.2141.21

,_

,_

,_

,_

,_

(0.5)a2.30 2591

,_

,_

31: = 30 5031

,_

,_

31: = 203

,_

(0.5)b. A = 31) 5 1 ( 3 . 2) 3 1 .( 3 . 220 . 6 3 . 26 . 2 9 . 48 108 108 8 109 4 5++(0.5)c. c1. 337= 0.(21) c2.227= 0,3(18) (0.5)c3. 0,(21) = 3379921 ; c4.5,1(6) = 561(0.5)Cu 2: (2)Gi khi lng ca 3 khi 7, 8, 9 ln lt l a, b, c (m3)a + b + c = 912 m3. (0.5)S hc sinh ca 3 khi l : 2 , 1a ;4 , 1b ;6 , 1cTheo ra ta c: 2 , 1 1 , 4 . 3a b v 6 , 1 . 5 4 , 1 . 4c b(0.5)206 , 1 . 15 4 , 1 . 12 2 , 1 . 4 c b a(0.5)Vy a = 96 m3 ;b = 336 m3 ;c = 480 m3.Nn s HS cc khi 7, 8, 9 ln lt l: 80 hs, 240 hs, 300 hs. (0.5)Cu 3: ( 1.5):a.Tm max A.Ta c: (x + 2)2 0 (x = 2)2 + 4 4 Amax= 43 khi x = -2 (0.75)b.Tm min B.Do (x 1)2 0 ; (y + 3)20 B1Vy Bmin= 1 khi x = 1 v y = -3 (0.75)Cu 4: (2.5)K CHct MB ti E. Ta c EAB cn ti E EAB =300 EAM = 200 CEA = MAE = 200 (0.5)Do ACB = 800 ACE = 400

AEC = 1200 ( 1 )(0.5)Mt khc: EBC = 200 v EBC = 400 CEB = 1200( 2 ) (0.5)Trang 43E300100MCB A HNm hc 2010-2011T ( 1 ) v ( 2 ) AEM = 1200Do EAC = EAM (g.c.g) AC = AM MAC cn ti A (0.5)V CAM = 400 AMC = 700. (0.5)Cu 5: (1.5)Gi s a2 v a + b khng nguyn t cng nhau a2 v a + bCng chia ht cho s nguyn t d: a2 chia ht cho d a chia htcho d v a + b chia ht cho d b chia hta cho d (0.5)(a,b) = d tri vi gi thit.Vy (a2,a + b) =1. (0.5)-------------------------------------------------------pn(ton7)Cu I :1) Xc nh a, b ,c654 32 1 + c b a=224 12 1020 9 5 4 3 524) 5 ( 412) 3 ( 310) 1 ( 5 + + c b a c b a=> a = -3 ; b = -11; c = -7.Cch 2 :654 32 1 + c b a = t ; sau rt a, b ,c thay vo tm t =- 2 tm a,b,c.2) Chng minht dcba = k => a= kb ; c = kdThay vo cc biu thc :03 25 33 25 33 25 3 23 25 3 22 222 222 2++ ++ ++ ++ kk kkk kcd dd cd cab bb ab a=> pcm.Cu II: Tnh:1) Ta c :2A= 2(99 . 971....7 . 515 . 31+ + +) = 993299131991971.....71515131 + + + =>A =99162) B = = 51 50 3 23131.....313131 + + + = ) 3 ( 1) 3 ( 1.....) 3 ( 1) 3 ( 1) 3 ( 151 50 3 2++ +++) 3 ( 1) 3 ( 1.....) 3 ( 1) 3 ( 1) 3 ( 152 51 4 3 2++++ => B31 ) 3 ( 13152= 525131 3 => B = 51513 . 4) 1 3 ( Cu IIITrang 44Nm hc 2010-2011Ta c : 0.2(3) = 0.2 + 0.0(3) = +102.1010,(1).3 =91.103102+ =3070,120(32) = 0,12 + 0,000(32) =0,12+10001.0,(32)= 0,12+10001.0,(01).32 =991.10003210012+=123751489Cu IV :Gi a thc bc hai l : P(x) = ax(x-1)(x-2) + bx(x-1)+c(x-3) + dP(0) = 10 => -3c+d =10 (1)P(1) = 12 => -2c+d =12 =>d =12+2c thay vo (1) ta c -3c+12+2c =10 =>c=2 , d =16P(2)= 4 => 2b -2+16 = 4 > b= -5P(3) = 1 => 6a-30 +16 =1 => a = 25Vy a thc cn tm l : P(x) =16 3 2 1 5 2 125+ + ) ( ) ( ) )( ( x x x x x x=> P(x) = 325x-10 122252+ + x xCu V:a) D thyADC =ABE ( c-g-c) => DC =BE .V AE AC; AD ABmt khc gc ADC = gc ABE=> DC Vi BE.b) Ta c MN // DC v MP // BE => MN MPMN = 21DC = 21BE =MP;Vy MNP vung cn ti M.---------------------------------------------------------p n 20Bi 1:a)A = 3 3 3 3 3 3 38 10 11 12 2 3 45 5 5 5 5 5 58 10 11 12 2 3 4 + + + + + + (0,25) Trang 45Nm hc 2010-2011A =1 1 1 1 1 1 13 38 10 11 12 2 3 41 1 1 1 1 1 15 58 10 11 12 2 3 4 _ _ + + + , ,+ _ _ + + + , , (0,25)A = 35 + 35 = 0 (0,25)b) 4B = 22 + 24 + ... + 2102(0,25) 3B = 2102 1;B = 1022 13 (0,25)Bi 2:a) Ta c 430 = 230.415(0,25)3.2410 = 230.311(0,25)m 415 > 311 430 > 311 230 + 330 + 430 > 3.2410(0,25)b) 4 = 36 >2933 >14 (0,25)36 +33 >29 +14 (0,25)Bi 3:Gi x1, x2 x3 ln lt l s ngy lm vic ca 3 my 1 2 33 4 5x x x (1) (0,25)Gi y1, y2, y3 ln lt l s gi lm vic ca cc my 1 2 36 7 8y y y (2) (0,25)Gi z1, z2, z3 ln lt l cng sut ca 3 my 5z1 = 4z2 = 3z3 1 2 31 1 15 4 3z z z (3) (0,25)Mx1y1z1 + x2y2z2 + x3y3z3 = 359 (3) (0,25)T (1) (2) (3) 1 1 1 2 2 2 3 3 33951518 40 39575 3 15x y z x y z x y z (0,5) x1y1z1 = 54; x2y2z2 = 105; x3y3z3 = 200 (0,25)Vy s thc mi i ln lt l 54, 105, 200 (0,25)Bi 4:a)EAB = CAD (c.g.c) (0,5)Trang 46Nm hc 2010-2011 ABM ADM (1) (0,25)Ta c + BMC MBD BDM (gc ngoi tam gic) (0,25) 0 0 060 60 120 BMC MBA BDM ADM BDM + + + + (0,25)b) Trn DM ly F sao cho MF = MB (0,5)FBM u (0,25)DFB AMB(c.g.c) (0,25) 0120 DFB AMB (0,5)Bi 6: Ta c12 (2) 3. ( ) 42x f f + (0,25)1 1 1( ) 3. (2)2 2 4x f f + (0,25) 47(2)32f (0,5)------------------------------------------------------- p n 21 Cu 1a.Nu x0 suy ra x = 1 (tho mn)Nu < 0 suy ra x = -3(tho mn)b. ' 6 316 32161xyx xy; hoc' 6 31xy;hoc23 3yx ' hoc33 2yx ' ;hoc63 1yx ' ; hoc63 1yx ' hoc23 3yx ' ;hoc33 2yx ' T ta c cc cp s (x,y) l (9,1); (-3, -1) ; (6, 2) ; (0,- 2) ; (5, 3) ; (1, -3) ; (4, 6); (2, -6)c. T 2x = 3y v 5x = 7z bin i v3 7 5 3 7 5 30221 14 10 61 89 50 63 89 50 15x y z x y z x y z + + x = 42; y = 28; z = 20Trang 47MABCDEFNm hc 2010-2011Cu 2a. A l tch ca 99 s m do 2 2 2 2 21 1 1 1 1.3 2.4 5.3 99.1011 1 1 .... 14 9 16 100 2 3 4 1001.2.3.2....98.99 3.4.5...99.100.101 101 1 12.3.4...99.100 2.3.4......99.100 200 2 2AA _ _ _ _ , , , , > < g g ggggb. B = 1 3 4 413 3 3x xx x x+ + + B nguyn( ) 44 33nguen xx U{ } 4; 25;16;1; 49 x Cu 3Thi gian i thc t nhiu hn thi gian d nhGi vn tc i d nh t C n B l v1 == 4km/hVn tc thc t i t C n B l V2 = 3km/hTa c: 1 1 12 2 24 33 4V t VvaV t V (t1 l thi gian i AB vi V1; t2 l thi gian i CB vi V2)t 1 2 1 2 123 15154 4 3 4 3 1t t t t tt t2 = 15 . 4 = 60 pht = 1 giVy qung ng CB l 3km, AB = 15kmNgi xut pht t 11 gi 45 pht (15:4) = 8 giCu 4a. Tam gic AIB = tam gic CID v c (IB = ID;gc I1 = gc I2; IA = IC)b. Tam gic AID = tam gic CIB (c.g.c) gc B1 = gc D1 v BC = AD hay MB =ND tam gic BMI = tam gic DNI (c.g.c) Gc I3 = gc I4 M, I, N thng hng v IM = INDo vy:I l trung im ca MNc. Tam gic AIB c gc BAI > 900 gc AIB < 900 gc BIC > 900d. Nu AC vung gc vi DC th AB vung gc vi AC do vy tam gic ABC vung ti ACu 5.P = 4 10 1014 4xx x + + P ln nht khi 104 x ln nhtTrang 48Nm hc 2010-2011Xt x > 4 th104 x < 0Xt x< 4 th104 x > 0104 x ln nht 4 x l s nguyn dng nh nht 4 x = 1 x = 3khi 104 x = 10 Pln nht = 11.-------------------------------------------------------------H ng dn chm 22 Bi 1 :a) Tm x . Ta c 6 2 x + 5x =96 2 x = 9-5x* 2x 6 0 x 3 khi 2x 6 = 9-5x x = 715khng tho mn. (0,5)* 2x 6 < 0x< 3 khi 6 2x = 9-5x x= 1tho mn. (0,5)Vy x = 1.b) Tnh . (1+2+3+...+90).( 12.34 6.68) :

,_

+ + +61514131 = 0. (0,5)( v 12.34 6.68 = 0).c) Ta c : 2A = 21 + 22 +23 + 24 + 25 +...+ 2101 2A A = 2101 1. (0,5)Nh vy 2101 1 < 2101 . Vy A1 . A = 5 tc l 4923511 +x xxx. (1)Bi 4 : E thuc phn gic ca ABCnn EN = EC ( tnh cht phn gic) suy ra :tam gic NEC cn v ENC = ECN (1) . D thuc phn gic ca gc CAB nn DC = DM(tnh cht phn gic ) suy ra tam gic MDC cn .v DMC =DCM,(2) . Ta li c MDB = DCM +DMC (gc ngoi ca CDM ) = 2DCM.Tng t ta li c AEN = 2ECN . M AEN = ABC (gc c cnh tng ng vung gc cng nhn).MDB = CAB (gc c cnh tng ng vung gc cng nhn ). Tam gic vung ABC cTrang 50Nm hc 2010-2011ACB = 900 , CAB + CBA = 900 , suy ra CAB = ABC = AEN + MDB = 2 ( ECN + MCD )suy raECN + MCD = 450 . Vy MCN = 900 450 =450 . (1,5)Bi 5 :Ta c P = -x2 8x + 5 = - x2 8x 16 +21 = -( x2 +8x + 16) + 21 = -( x+ 4)2 + 21; (0,75)Do ( x+ 4)20 vi mi x nn ( x +4)2 +21 21 vi mi x . Du (=) xy ra khi x = -4Khi P c gi tr ln nht l 21.------------------------------------------------------------h ng dn 23 Cu 1: (3)b/ 2-1.2n + 4.2n = 9.25suy ra 2n-1 + 2n+2 = 9.25 0,5suy ra 2n (1/2 +4) = 9. 25suy ra 2n-1 .9 =9. 25 suy ra n-1 = 5 suy ra n=6. 0,5c/ 3n+2-2n+2+3n-2n=3n(32+1)-2n(22+1) = 3n.10-2n.5 0,5v 3n.10 M10 v 2n.5 = 2n-1.10 M10 suy ra 3n.10-2n.5 M10 0,5Bi 2:a/ Gi x, y, z ln lt l s hc sinh ca 7A, 7B, 7C tham gia trng cy(x, y, zz+) ta c: 2x=3y = 4z v x+y+z =130 0,5hay x/12 = y/8 = z/6 m x+y+z =130 0,5suy ra: x=60; y = 40; z=30-7(4343-1717)b/ -0,7(4343-1717) = 0,510Ta c: 4343 = 4340.433= (434)10.433 v 434 tn cng l 1 cn 433 tn cng l 7 suy ra 4343 tn cng bi 7Trang 51Nm hc 2010-20111717 = 1716.17 =(174)4.17v 174 c tn cng l 1 suy ra (174)4 c tn cng l 1 suy ra 1717 = 1716.17 tn cng bi 7 0,5suy ra 4343 v 1717 u c tn cng l 7 nn 4343-1717 c tn cng l 0 suy ra 4343-1717 chia ht cho 10 0,5suy ra-0,7(4343-1717) l mt s nguyn.Bi 3: 4( Hc sinh t v hnh)a/ MDB= NECsuy ra DN=EN0,5b/ MDI= NEI suy ra IM=IN suy ra BC ct MN ti im I l trung im ca MN0,5c/ Gi H l chn ng cao vung gc k t A xung BC ta c AHB= AHC suy ra HAB=HAC0,5gi O l giao AH vi ng thng vung gc vi MN k t I th OAB= OAC (c.g.c) nn OBA = OCA(1)0,5 OIM= OIN suy ra OM=ON 0,5suy ra OBN= OCN (c.c.c) OBM=OCM(2)0,5T (1) v (2) suy ra OCA=OCN=900 suy ra OC AC 0,5Vy im O c nh.-------------------------------------------------------p n 24Cu 1: (2).a. |a| + a = 2a vi a 0 (0,25)Vi a < 0 th |a| + a = 0 (0,25).b. |a| - a-Vi a0 th |a| - a = a a = 0-Vi a< 0 th |a| - a = - a - a = - 2ac.3(x 1) - 2|x + 3|-Vi x + 3 0 x - 3Ta c: 3(x 1) 2 |x + 3| = 3(x 1) 2(x + 3)= 3x 3 2x 6= x 9.(0,5)-Vi x + 3 < 0 x< - 3Tac: 3(x 1) - 2|x + 3| = 3(x 1) + 2(x + 3).Trang 52Nm hc 2010-2011= 3x 3 + 2x + 6= 5x + 3 (0,5).Cu 2: Tm x(2).a.Tm x,bit: |5x - 3| - x = 7 5 3 7 x x + (1)(0,25 )K:x -7(0,25 )( )( )5 3 715 3 7x xx x + +.(0,25 )Vy c hai gi tr x tha mn iu kin u bi. x1 = 5/2;x2= - 2/3 (0,25).b. |2x + 3| - 4x< 9 (1,5) |2x + 3| < 9 + 4x(1)K:4x +9 0 x 94(1)( ) 4 9 2 3 4 9 x x x + < < + 2 3 x < < (t/mK)(0,5).Cu 3:Gi ch s ca s cn tm l a, b, c. V s cn tm chia ht 18 s phi chia ht cho 9.Vy (a + b + c ) chia ht cho 9. (1)(0,5).Tac: 1 a + b + c 27 (2)V1 a 9 ; b 0 ; 0 c 9T (1) v (2) ta c (a + b + c) nhn cc gi tr 9, 18, 27(3).Suy ra:a = 3 ; b = 6 ; c = 9(0,5).V s cn tm chia ht 18 nn va chia ht cho 9 va chia ht cho 2 ch s hng n v phi l s chn.Vy ss cn tm l: 396 ; 963(0,5). -V hnh ng vit gi thit, kt lun ng(0,5).-Qua N k NK // AB ta c.EN // BK NK = EBEB // NKEN = BKLi c: AD = BE (gt) AD = NK (1)-Hc sinh chng minh ADM = NKC (gcg) (1) DM= KC (1)------------------------------------------------------p n 25Trang 53Nm hc 2010-2011Bi 1:Ta c:10A = 20072007 200710 10 9=1+10 1 10 1++ +(1)Tng t: 10B = 20082008 200810 10 9=1+10 1 10 1++ +(2)T (1) v (2) ta thy : 2007 20089 910 1 10 1>+ + 10A > 10BA > BBi 2:(2im) Thc hin php tnh:A = 1 1 11 . 1 ... 1(1 2).2 (1 3).3 (1 2006)20062 2 2 _ _ _ + + + , , , = 2 59 2007.2006 2 4 10 18 2007.2006 2. . .... . . ....3 6 10 2006.2007 6 12 20 2006.2007 (1)M: 2007.2006 - 2 = 2006(2008 - 1) + 2006 - 2008= 2006(2008 - 1+ 1) - 2008 = 2008(2006 -1) = 2008.2005 (2)T (1) v (2) ta c:A =4.1 5.2 6.3 2008.2005 (4.5.6...2008)(1.2.3...2005) 2008 1004. . ....2.3 3.4 4.5 2006.2007 (2.3.4...2006)(3.4.5...2007) 2006.3 3009 Bi 3:(2im) T:x 1 1 1 x 18 y 4 y 8 4 Quy ng mu v phi ta c :1 x-2y 8. Do : y(x-2) =8. x, y nguyn th y v x-2 phi l c ca 8. Ta c cc s nguyn tng ng cn tm trong bng sau:Y 1 -1 2 -2 4 -4 8 -8x-2 8 -8 4 -4 2 -2 1 -1X 10 -6 6 -2 4 0 3 1Bi 4:(2 im)Trong tam gic tng di hai cnh ln hn cnh th 3. Vy c:b + c > a.Nhn 2 v vi a >0 ta c: a.b + a.c > a2. (1)Tng t ta c : b.c + b.a > b2(2)a.c + c.b > c2(3).Cng v vi v ca (1), (2), (3) ta c:Trang 54Nm hc 2010-20112(ab + bc + ca) > a2 + b2 + c2.Bi 5:(3 im)V tia phn gic ABK ct ng thng CK I.Ta c:IBC Vcn nn IB = IC.BIA V=CIA V (ccc) nn 0BIACIA 120 . Do :BIA V= BIK V(gcg)BA=BK b) T chng minh trn ta c:0BAK 70 ---------------------------------------------------p n 26Cu 1: ( 2 im )a. Do 11 12 2+kkk vi k = 1,2..n ( 0,25 im )p dng bt ng thc C Si cho k +1 s ta c:Trang 55CKAIBNm hc 2010-2011( ) 11111 111 ... 1 11.. 1 .... 1 . 1 11 1++ +++++ + + + c b ahchbha1 1 1 (0 , 4 im )=> a :b : c = 6 : 15 : 1051:21:31 1:1:1 c b ah h h(0 ,4 im )Vy a: b: c = 10 : 10 : 6Cu 4: ( 2 im )Trn tia Ox ly A , trn tia Oy lyBsao cho O A= OB= a( 0,25 im )Ta c: O A+ OB= OA + OB = 2a=> A A= BB ( 0,25 im )GiH v K ln lt l hnh chiuCa A v Btrn ng thngA BTam gic HA A= tam gic KBB( cnh huyn, gc nhn ) ( 0,5 im )=> H, B K A do HK =B A (0,25 im)Trang 56yNm hc 2010-2011Ta chng minh cHKAB (Du = A trngA BtrngB (0,25 im)do AB B A ( 0,2 im )Vy AB nh nht OA = OB = a(0,25im )Cu 5 ( 2 im )Gi sQ d c b a + +( 0,2 im )=>a d b a +=> b +b+2a d a d bc 22+ + ( 0,2 im)=> 2 ( ) a d c b a d bc 22 + ( 1 ) ( 0,2 im)=> 4bc = ( ) c b a d +2 2 + 4 d2a 4b ( ) c b a d +2a ( 0,2 im)=> 4 d ( ) c b a d +2a=( ) c b a d +2 2 + 4d 2a 4 bc ( 0,2 im)* Nu 4 d ( ) c b a d +2 # 0 th:( )) ( 44 422 2 2c b a d dab a d c b a da + + +l s hu t (0,2 5im )** Nu 4 d ( ) c b a d +2= 0 th: d =0 hoc d 2+ a-b c = 0 ( 0,25 im )+ d = 0 ta c : 0 + + c b a=> Q c b a 0(0,25 im )+ d 2+ a-b c = 0 th t(1 ) =>a d bc V a, b, c, d 0 nn Q a 0( 0,25 im )Vyal s hu t.Do a,b,c c vai tr nh nhaunn c b a , , l cc s hu t--------------------------------------------------Trang 57Nm hc 2010-2011 1 Bi 1. (4 im)a) Chng minh rng 76 + 75 74 chia ht cho 55b) Tnh A = 1 + 5 + 52 + 53 + . . . + 549 + 55 0Bi 2. (4 im)a) Tm cc s a, b, c bit rng : 2 3 4a b c v a + 2b 3c = -20b) C 16 t giy bc loi 20 000, 50 000, 100 000. Tr gi mi loi tin trn u bng nhau. Hi mi loi c my t?Bi 3. (4 im)a) Cho hai a thc f(x) = x5 3x2 + 7x4 9x3 + x2 - 14x g(x) = 5x4 x5 + x2 2x3 + 3x2 - 14Tnh f(x) + g(x) v f(x) g(x).b) Tnh gi tr ca a thc sau: A = x2 + x4 + x6 + x8 + + x100ti x = -1.Bi 4. (4 im)Cho tam gic ABC c gc A bng 900, trn cnh BC ly im E sao cho BE = BA. Tia phn gic ca gc B ct AC D.a) So snh cc di DA v DE.Trang 58Nm hc 2010-2011b) Tnh s o gc BED.Bi 5. (4 im)Cho tam gic ABC, ng trung tuyn AD. K ng trung tuyn BE ct AD G. Gi I, K theo th t l trung im ca GA, GB. Chng minh rng:a) IK// DE, IK = DE.b) AG = 23AD. 2: Mn: Ton 7Bi 1: (3 im): Tnh1 1 2 2 318 (0, 06: 7 3 .0, 38) : 19 2 .46 2 5 3 4 1 _ + 1 ] , Bi 2: (4 im):Choa cc bchngminhrng:a)2 22 2a c ab c b++b)2 22 2b a b aa c a +Bi 3:(4 im)Tmxbit:a) 14 25x + b) 15 3 6 112 7 5 2x x + Bi 4:(3 im)Mtvtchuynngtrncccnhhnhvung.Trnhaicnhuvtchuynngvivntc5m/s,trncnhthbavivntc4m/s,trncnhthtvivntc3m/s.Hidicnhhnhvungbitrngtngthigianvtchuynngtrnbncnhl59giyBi 5:(4 im)ChotamgicABCcntiAc0A 20 ,vtamgicuDBC(DnmtrongtamgicABC).TiaphngiccagcABDctACtiM.Chngminh:Trang 59Nm hc 2010-2011a) TiaADlphngiccagcBACb) AM=BCBi 6:(2 im):Tm, xy bit:2 225 8( 2009) y x 3Bi 1:(4 im)a)Thchinphptnh:( )( )12 5 6 2 10 3 5 26 39 32 4 52 .3 4 .9 5 .7 25 .49A125.7 5 .142 .3 8 .3 ++b)Chngminhrng:Vimisnguyndngnth:2 23 2 3 2n n n n + + + chiahtcho10Bi 2:(4 im)Tmxbit:a. ( )1 4 23, 23 5 5x + +b. ( ) ( )1 117 7 0x xx x+ + Bi 3: (4 im)a) SAcchiathnh3stltheo23 1: :54 6 .Bitrngtngccbnhphngcabasbng24309.TmsA.b) Choa cc b.Chngminhrng:2 22 2a c ab c b++Bi 4: (4 im)ChotamgicABC,MltrungimcaBC.TrntiaicacatiaMAlyimEsaochoME=MA.Chngminhrng:a)AC=EBv AC//BEb)GiIlmtimtrnAC;KlmtimtrnEBsaochoAI=EK.ChngminhbaimI,M,Kthnghngc)TEk EH BC ( ) H BC .BitHBE=50o;MEB=25o.TnhHEM vBMEBi 5: (4 im)ChotamgicABCcntiAc0A 20 ,vtamgicuDBC(DnmtrongtamgicABC).TiaphngiccagcABDctACtiM.Chngminh:Trang 60Nm hc 2010-2011c) TiaADlphngiccagcBACd) AM=BCTrang 61Nm hc 2010-2011 4 Bi 1: (2 im)Cho A = 2-5+8-11+14-17++98-101a, Vit dng tng qut dng th n ca Ab, Tnh ABi 2: ( 3 im) Tm x,y,ztrong cc trng hp sau:a, 2x = 3y =5z v2 x y =5b, 5x = 2y, 2x = 3z v xy = 90.c, 1 2 3 1 y z x z x yx y z x y z+ + + + + + +Bi 3: ( 1 im)1. Cho 3 8 9 1 22 3 4 9 1...a a a a aa a a a a v (a1+a2++a9 0) Chng minh: a1 = a2 = a3== a92. Cho t l thc: a b c a b ca b c a b c+ + ++ v b 0 Chng minh c = 0Bi 4: ( 2 im) Cho 5 s nguyn a1, a2, a3, a4, a5. Gi b1, b2, b3, b4, b5 l hon v ca 5 s cho. Chng minh rng tch (a1-b1).(a2-b2).(a3-b3).(a4-b4).(a5-b5) M2Bi 5: ( 2 im)Cho on thng AB v O l trung im ca on thng . Trn hai na mt phng i nhau qua AB, k hai tia Ax v By song song vi nhau. Trn tia Ax ly hai im D v F sao cho AC = BD v AE = BF. Chng minh rng : ED = CF.Trang 62Nm hc 2010-2011=== Ht=== 5 Bi 1: (3 im)1. Thc hin php tnh: 14, 5: 47, 375 26 18.0, 75 .2, 4: 0, 8832 517, 81:1, 37 23 :13 6 1 _ 1 , ]2.Tm cc gi tr ca x v y tho mn:( )2007 20082 27 3 10 0 x y + + 3.Tm cc s a, b sao cho2007abl bnh phng ca s t nhin.Bi 2: ( 2 im)1. Tm x,y,z bit: 1 2 32 3 4x y z v x-2y+3z = -102. Cho bn s a,b,c,d khc 0 v tho mn: b2 = ac; c2 = bd; b3 + c3 + d3 0Chng minh rng: 3 3 33 3 3a b c ab c d d+ ++ +Bi 3: ( 2 im)1. Chng minh rng: 1 1 1 1... 101 2 3 100+ + + + >2. Tm x,y C = -18- 2 6 3 9 x y + t gi tr ln nht.Bi 4: ( 3 im)Cho tam gic ABC vung cn ti A c trung tuyn AM. E l im thuc cnh BC.K BH, CK vung gc vi AE (H, K thuc AE). 1, Chng minh: BH = AK 2, Cho bit MHK l tam gic g? Ti sao?Trang 63Nm hc 2010-2011=== Ht=== s 6Cu 1: Tm cc s a,b,c bit rng: ab =c ;bc= 4a; ac=9bCu 2:Tm s nguyn x tho mn:a, 5x-3< 2 b, 3x+1>4 c,4- x+2x =3Cu3: Tm gi tr nh nht ca biu thc:A = x+ 8 -xCu 4: Bit rng :12+22+33+...+102= 385.Tnh tng : S= 22+ 42+...+202Cu 5 :Cho tam gic ABC ,trung tuyn AM .Gi I l trung im ca on thng AM, BI ct cnh AC ti D.a. Chng minh AC=3 ADb. Chng minh ID =1/4BD------------------------------------------------- Ht ------------------------------------------ s 7Thi gian lm bi: 120 phtCu 1 . ( 2) Cho: dccbba . Chng minh:dad c bc b a ,_+ ++ +3.Cu 2. (1). Tm A bit rng: A = a c bb a cc b a+++.Cu 3. (2). Tm Z x A Z v tm gi tr .a). A = 23+xx.b). A =32 1+xx.Trang 64Nm hc 2010-2011Cu 4.(2).Tm x, bit:a)3 x = 5 . b).( x+ 2) 2 = 81.c). 5 x + 5 x+ 2 = 650Cu 5.(3). Cho ABC vung cn ti A, trung tuyn AM . E BC, BH AE, CK AE, (H,K AE). Chng minh MHK vung cn.-------------------------------- Ht ------------------------------------ s 8Thi gian lm bi : 120 pht.Cu 1 : ( 3 im).1. Ba ng cao ca tam gic ABCc di l 4,12 ,a . Bit rng a l mt s t nhin. Tm a ?2. Chng minh rng t t l thc dcba ( a,b,c ,d0, a b, c d) ta suy ra c cc t l thc:a) d c cb a a. b) dd cb b a ++.Cu 2: ( 1 im). Tm s nguyn x sao cho: ( x2 1)( x2 4)( x2 7)(x2 10) < 0.Cu 3: (2 im).Tm gi tr nh nht ca:A = x-a+ x-b+ x-c+ x-dvi a 1. c.2 3 x + 5.Cu2: ( 2 im)a. Tnh tng: A= (- 7) + (-7)2 + + (- 7)2006 + (- 7)2007. Chng minh rng: A chia ht cho 43.b. Chng minh rng iu kin cn v m2 + m.n + n2 chia ht cho 9 l: m, n chia ht cho 3.Cu 3: ( 23,5 im) di cc cnh ca mt tam gic t l vi nhau nh th no,bit nu cng ln lt di tng hai ng cao ca tam gic th cc tng ny t l theo 3:4:5.Trang 71Nm hc 2010-2011Cu 4: ( 3 im ) Cho tam gic ABC cn ti A. D l mt im nm trong tam gic, bitADB> ADC . Chng minh rng: DB < DC.Cu 5: ( 1 im ) Tm GTLN ca biu thc: A = 1004 x - 1003 x +.-------------------------------------- Ht --------------------------------- s 18Cu 1 (2 im): Tm x, bit : a. 3x 2 +5x = 4x-10 b. 3+ 2x 5 +> 13Cu 2: (3 im )a.Tm mt s c 3 ch s bit rng s chia ht cho 18 v cc ch s ca n t l vi 1, 2, 3.b. Chng minh rng:Tng A=7 +72+73+74+...+74n chia ht cho 400(nN). Cu 3 : (1im )cho hnh v , bit ++ = 1800chng minh Ax// By. A xC B yCu 4 (3 im ) Cho tam gic cn ABC, c ABC=1000. K phn gic trong ca gc CAB ct AB ti D. Chng minh rng:AD + DC =ABCu 5 (1 im )Tnh tng.S = (-3)0 + (-3)1+ (-3)2 + .....+ (-3)2004.------------------------------------ Ht ---------------------------------- s 19Thi gian lm bi: 120 phBi 1: (2,5) Thc hin php tnh sau mt cch hp l:1 1 1 1 1 1 1 1 190 72 56 42 30 20 12 6 2 Bi 2: (2,5)Tnh gi tr nh nht ca biu thc: A = x x + 5 2Trang 72Nm hc 2010-2011Bi 3: (4) Cho tam gic ABC. Gi H, G,O ln lt l trc tm , trng tm v giao im ca 3 ngtrung trc trong tam gic. Chng minh rng:a. AH bng 2 ln khong cch t O n BCb. Ba im H,G,O thng hng v GH= 2 GOBi 4: (1 ) Tm tng cc h s ca a thc nhn c sau khi b du ngoc trong biu thc (3-4x+x2)2006.(3+ 4x + x2)2007.------------------------------------------- Ht ------------------------------------------ 20Thi gian lm bi: 120 phtCu 1(3): Chng minh rngA = 22011969 + 11969220 + 69220119 chia ht cho 102Cu 2(3): Tm x, bit: a. x x 2 3 + + ; b.3x 5 x 2 +Cu 3(3): Cho tam gic ABC. Gi M, N, P theo th t l trung im ca BC, CA, AB. Cc ng trung trc ca tam gic gp nhau tai 0. Cc ng cao AD, BE, CF gp nhau ti H. Gi I, K, R theo th t l trung im ca HA, HB, HC.a) C/m H0 v IM ct nhau ti Q l trung im ca mi on.b) C/m QI = QM = QD = 0A/2c) Hy suy ra cc kt qu tng t nh kt qu cu b.Cu 4(1):Tm gi tr ca x biu thcA = 10 - 3|x-5|t gi tr ln nht.--------------------------------------------- Ht --------------------------------------------- 21: Trang 73Nm hc 2010-2011Bi 1: (2) Cho biu thcA = 35+xxa) Tnh gi tr ca A ti x = 41b) Tm gi tr ca x A = - 1c) Tm gi tr nguyn ca x A nhn gi tr nguyn.Bi 2. (3)a) Tm x bit:1 7 x xb) Tnh tng M = 1 + (- 2) + (- 2)2 + +(- 2)2006c) Cho a thc: f(x) = 5x3 + 2x4 x2 + 3x2 x3 x4 + 1 4x3. Chng t rng a thc trn khng c nghimBi 3.(1Hi tam gic ABC l tam gic gbit rng cc gc ca tam gic t l vi 1, 2, 3.Bi 4.(3) Cho tam gic ABC c gc B bng 600. Hai tia phn gic AM v CN ca tam gic ABC ct nhau ti I.a) Tnh gc AICb) Chng minh IM = INBi 5. (1)Cho biu thcA = x x62006.Tm gi tr nguyn ca x A t gi tr ln nht. Tm gi tr ln nht .---------------------------------------- Ht -------------------------------------- 22Cu 1:1.Tnh:a. 20 1521,_,_41. b. 30 2591,_,_31:2. Rt gn:A = 20 . 6 3 . 26 . 2 9 . 48 8 109 4 5+3. Biu din s thp phn di dng phn s v ngc li:a. 337b. 227c. 0, (21) d. 0,5(16)Cu 2:Trong mt t lao ng, ba khi 7, 8, 9 chuyn ch c 912 m3 t. Trung bnh mi hc sinh khi 7, 8, 9 theo th tlm -Trang 74Nm hc 2010-2011c 1,2 ; 1,4 ; 1,6 m3 t. S hc sinh khi 7, 8 t l vi 1 v 3. Khi 8 v 9 t l vi 4 v 5. Tnh s hc sinh mi khi.Cu 3:a.Tm gi tr ln nht ca biu thc:A = 4 ) 2 (32+ + xb.Tm gi tr nh nht ca biu thc:B = (x+1)2 + (y + 3)2 + 1Cu 4: Cho tam gic ABC cn (CA = CB) v C = 800. Trong tam gic sao cho 0MBA 30 v010 MAB .TnhMAC.Cu 5:Chng minh rng : nu (a,b) = 1 th (a2,a+b) = 1.------------------------------------- Ht -------------------------------------23Thi gian: 120 pht.Cu I: (2)1) Cho 654 32 1 + c b a v 5a - 3b - 4 c = 46 . Xc nh a, b, c2) Cho t l thc : dcba . Chng minh : cd dd cd cab bb ab a3 25 3 23 25 3 222 222 2++ ++ . Vi iu kin mu thc xc nh.Cu II : Tnh : (2)1) A = 99 . 971....7 . 515 . 31+ + +2) B = 51 50 3 23131.....313131 + + + Cu III : (1,5 )i thnh phn s cc s thp phn sau :a. 0,2(3) ;b.1,12(32).Cu IV : (1.5) Xc nh cc a thc bc 3 bit :P(0) = 10; P(1) = 12; P(2) = 4 ; p(3) = 1Cu V : (3)Cho tam gic ABC c 3 gc nhn. Dng ra pha ngoi 2 tam gic vung cn nh A l ABD v ACE . Gi M;N;P ln lt l trung im ca BC; BD;CE .a. Chng minh : BE = CD v BE vi CDb. Chng minh tam gic MNP vung cn---------------------------------------------- Ht -----------------------------------------------Trang 75Nm hc 2010-2011 24Thi gian lm bi: 120 phtBi 1 (1,5): Thc hin php tnh:a) A = 3 30,375 0,31,5 1 0,7511 125 5 50,265 0,5 2,5 1,2511 12 3 + ++ + + + b) B = 1 + 22 + 24 + ... + 2100Bi 2 (1,5):a) So snh: 230 + 330 + 430 v 3.2410b) So snh: 4 +33 v29+ 14Bi 3 (2):Ba my xay xay c 359 tn thc. S ngy lm vic ca cc my t l vi 3:4:5, s gi lm vic ca cc my t l vi 6, 7, 8, cng sut cc my t l nghc vi 5,4,3. Hi mi my xay c bao nhiu tn thc.Bi 4 (1):Tm x, y bit:a) 3 4 x 3 b) 1 1 1 1... 21.2 2.3 99.100 2x _+ + + ,Bi 5 ( 3):ChoABC c cc gc nh hn 1200. V pha ngoi tam gic ABC cc tam gic u ABD, ACE. Gi M l giao im ca DC v BE. Chng minh rng:a) 0120 BMC b) 0120 AMBBi 6 (1):Cho hm s f(x) xc nh vi mi x thuc R. Bit rng vi mi x ta u c:21( ) 3. ( ) f x f xx+ . Tnh f(2).---------------------------------------- Ht ------------------------------------------Trang 76Nm hc 2010-2011 25Thi gian lm bi: 120 phtCu 1 (2) Tm x, y, z Z, bita. x x + = 3 - xb.21 16 yxc. 2x = 3y; 5x = 7z v 3x - 7y + 5z = 30Cu 2 (2)a. Cho A =) 11001)...( 141).( 131).( 121(2 2 2 2 .Hy so snh A vi21b. Cho B = 31+xx .Tm x Z B c gi tr l mt s nguyn dngCu 3 (2)Mt ngi i t A n B vi vn tc 4km/h v d nh n B lc 11 gi 45 pht. Sau khi i c 51 qung ng th ngi i vi vn tc 3km/h nn n B lc 12 gi tra.Tnh qung ngAB v ngi khi hnh lc my gi?Cu 4 (3) Cho ABC c A > 900. Gi I l trung im ca cnh AC. Trn tia i ca tia IB ly im D sao cho IB = ID. Ni c vi D.a. Chng minh CID AIB b. Gi M l trung im ca BC; N l trung im ca CD. Chng minh rng I l trung im ca MNc. Chng minh AIB AIB BIC a2 + b2 + c2.Bi 5:(3 im) Cho tam gic ABC c 0B=C=50. Gi K l im trong tam gic sao cho 0 0KBC=10 KCB=30a. Chng minh BA = BK.b. Tnh s o gc BAK.--------------------------------- Ht ----------------------------------Trang 81Nm hc 2010-2011 thi 30Thi gian lm bi: 120 phtCu 1.Vi mi s t nhin n 2hy so snh:a. A= 2 2 2 21....413121n+ + + + vi 1 .b. B = ( )2 2 2 221...614121n+ + + + vi 1/2Cu 2:Tm phn nguyn ca, vi 1 4 31....34232+++ + + + nnnCu 3:Tm t l 3 cnh ca mt tam gic, bit rng cng ln lt di hai ng cao ca tam gic th t l cc kt qu l5: 7 : 8.Cu 4:Cho gc xoy , trn hai cnh ox v oy ln lt ly cc im A v B cho AB c di nh nht.Cu 5:Chng minh rng nu a, b, c vc b a + +l cc s hu t.--------------------------------------------------------------Trang 82Nm hc 2010-2011p n - 1 Bi 1. 4a) 74( 72 + 7 1) = 74. 55 M55 (pcm)2b) Tnh A = 1 + 5 + 52 + 53 + . . . + 549 + 55 0(1) 5.A = 5 + 52 + 53 + . . . + 549 + 55 0 + 551 (2)1Tr v theo v (2) cho (1) ta c : 4A = 551 1 => A = 511451Bi 2. 4a) 2 3 4a b c 2 3 2 3 2052 6 12 2 6 12 4a b c a b c + + => a = 10, b = 15, c =20.2b) Gi s t giy bc 20 000, 50 000, 100 000 theo th t l x, y, z ( x, y, z N*) 0,5Theo bi ra ta c: x + y + z = 16v 20 000x = 50 000y = 100 000z0,5Bin i: 20 000x = 50 000y = 100 000z=> 20000 50000 100000 162100000 100000 100000 5 2 1 5 2 1 8x y z x y z x y z + + + + 0,5Suy ra x = 10, y = 4, z = 2.Vy s t giy bc loi 20 000, 50 000, 100 000 theo th t l 10; 4; 2.0,5Bi 3. 4Trang 83Nm hc 2010-2011a) f(x) + g(x) = 12x4 11x3 +2x2- 14x - 14 1 f(x) - g(x)= 2x5 +2x4 7x3 6x2 - 14x + 14 1b) A = x2 + x4 + x6 + x8 + + x100 ti x = - 1A = (-1)2 + (-1)4 + (-1)6 ++ (-1)100 = 1 + 1 + 1 ++ 1 = 50 (c 50 s hng)2Bi 4. 4: V hnh (0,5) phn a) 1,5 - phn b) 2a)ABD =EBD (c.g.c) => DA = DEb) V ABD =EBD nn gc A bng gc BEDDo gc A bng 900 nn gc BED bng 900 edcabBi 5: 4a) Tam gic ABC v tam gic ABG c:DE//AB, DE = 12AB, IK//AB, IK= 12ABDo DE // IK v DE = IKb) GDE =GIK (g. c. g) v c: DE = IK (cu a)Gc GDE = gc GIK (so le trong, DE//IK)Gc GED = gc GKI (so le trong, DE//IK) GD = GI. Ta c GD = GI = IA nn AG = 23ADGkiedcba- V hnh: 0,5- Phn a) ng: 2- Phn b) ng: 1,5 2: Bi 1: 3 im1 1 2 2 318 (0, 06: 7 3 .0, 38) : 19 2 .46 2 5 3 4 1 _ + 1 ] , ==109 6 15 17 38 8 19( : . ) : 19 .6 100 2 5 100 3 4 1 _ + 1 ] , 0.5=109 3 2 17 19 38. . : 196 50 15 550 3 1 _ _ + 1 , , ]1Trang 84Nm hc 2010-2011=109 2 323 19:6 250 250 3 1 _ + 1 , ]0.5= 109 13 3.6 10 19 _ ,= 0.5=5063 253.30 19 950.5Bi 2:a) Ta cc bsuyra2. c a b 0.5khi2 2 22 2 2..a c a a bb c b a b+ ++ +0.5 =( )( )aa b ab a b b++0.5b)Theocua)tac:2 2 2 22 2 2 2a c a b c bb c b a c a+ + + +0.5t2 2 2 22 2 2 21 1b c b b c ba c a a c a+ + + +1 hay2 2 2 22 2b c a c b aa c a+ + 0.5vy2 22 2b a b aa c a +0.5Bi 3: a)14 25x + 12 45x + +0.51 12 25 5x x + + hoc125x + 1Vi1 12 25 5x x + hay95x 0.25Vi1 12 25 5x x + hay115x 0.25b)15 3 6 112 7 5 2x x + 6 5 3 15 4 7 2x x + +0.5Trang 85Nm hc 2010-20116 5 13( )5 4 14x + 0.549 1320 14x 0.5130343x 0.5Bi 4: Cngmtonng,cntcvthigianlhaiilngtlnghch0.5Gix,y,zlthigianchuynnglnltviccvntc5m/s;4m/s;3m/sTac:5. 4. 3. x y z v59 x x y z + + + 1hay:59601 1 1 1 1 1 1 595 4 3 5 5 4 3 60x y z x x y z + + + + + +0.5Do:160. 125x ;160. 154x ;160. 203x 0.5Vycnhhnhvungl:5.12=60(m)0.5Bi 5: -Vhnh,ghiGT,KLng 0.5a)Chngminh ADB= ADC(c.c.c) 1suyra DAB DAC Do0 020 : 2 10 DAB b) ABC cn ti A, m 020 A (gt) nn0 0 0(180 20 ) : 2 80 ABC ABCunn060 DBC Tia BD nm gia hai tia BA v BC suy ra0 0 080 60 20 ABD .TiaBMlphngiccagcABDnn010 ABM XttamgicABMvBADc:ABcnhchung; 0 020 ; 10 BAM ABD ABM DAB Vy: ABM= BAD(g.c.g)suyraAM=BD,mBD=BC(gt)nnAM=BCBi 6: 2 225 y 8(x 2009) Tac8(x-2009)2=25-y28(x-2009)2+y2=25(*)0.5Trang 86200MABCDNm hc 2010-2011Vy2 0nn(x-2009)2258,suyra(x-2009)2=0hoc(x-2009)2=1 0.5Vi(x-2009)2=1thayvo(*)tacy2=17(loi)Vi(x-2009)2=0thayvo(*)tacy2=25suyray=5(doy )0.5Ttmc(x=2009;y=5) 0.5----------------------------------------------------------------------- 3Bi 1:(4im):pnThangima)(2im)( )( )( )( )( )( )( )1012 5 6 2 10 3 5 2 12 5 12 4 10 3 46 3 12 6 12 5 9 3 9 3 39 32 4 512 4 10 312 5 9 3 310 3 12 412 5 9 32 .3 4 .9 5 .7 25 .49 2 .3 2 .3 5 .7 5 .72 .3 2 .3 5 .7 5 .2 .7125.7 5 .142 .3 8 .32 .3 . 3 1 5 .7 . 1 72 .3 . 3 1 5 .7 . 1 25 .7 . 6 2 .3 .22 .3 .4 5 .7 .91 10 76 3 2A + +++ + + b)(2im)3n+2-Vimisnguyndngntac:2 23 2 3 2n n n n + + + =2 23 3 2 2n n n n + ++ =2 23 (3 1) 2 (2 1)n n+ +=13 10 2 5 3 10 2 10n n n n 0,5im0,5im0,5im0,5im0,5im1imTrang 87Nm hc 2010-2011=10(3n-2n)Vy2 23 2 3 2n n n n + + + M10viminlsnguyndng. 0,5imBi 2:(4im)pnThangima)(2im)( )1 231231 723 31 523 31 4 2 1 4 16 23, 23 5 5 3 5 5 51 4 143 5 5123xxxxx xxx+ + + + + + + b)(2im)( ) ( )( ) ( )1 111 107 7 07 1 7 0x xxx xx x+ ++ 1 ]( )( )( )1 101107 01 ( 7) 07 0 7( 7) 1 87 1 7 010xxxxx xx xx x+ _ ,+ 1 ] 0,5im0,5im0,5im0,5im0,5im0,5im0,5im0,5imBi 3: (4im)pn Thangima)(2,5im)Gia,b,clbascchiaratsA.Theobitac:a:b:c=231: :546 (1)0,5imTrang 88Nm hc 2010-2011va2+b2+c2=24309(2)T(1) 2 3 15 4 6a b c =k2 3; ;5 4 6ka k b kc Do(2)24 9 1( ) 2430925 16 36k + + k=180vk= 180 +Vik=180,tac:a=72;b=135;c=30.KhitacsA=a+b+c=237.+Vik= 180 ,tac:a= 72 ;b= 135 ;c= 30 KhitacsA= 72 +( 135 )+(30 )= 237 .b)(1,5im)Ta cc bsuyra2. c a b khi2 2 22 2 2..a c a a bb c b a b+ ++ +=( )( )a a b ab a b b++0,5im0,5im0,5im0,5im0,5im0,5im0,5imBi 4:(4im)pnThangimVhnh 0,5ima/(1im)Xt AMC v EMB c:AM=EM(gt)Trang 89KHEMBACINm hc 2010-2011AMC =EMB(inh)BM=MC(gt)Nn: AMC = EMB (c.g.c) 0,5imAC=EBV AMC = EMB MAC =MEB(2gccvtrsoletrongctobingthngACvEBctngthngAE)SuyraAC//BE. 0,5imb/(1im)Xt AMI v EMK c:AM=EM(gt)MAI =MEK (v AMC EMB )AI=EK(gt)Nn AMI EMK (c.g.c) 0,5 im Suy raAMI =EMK MAMI +IME=180o(tnhchthaigckb)EMK+IME=180oBaimI;M;Kthnghng 0,5imc/(1,5im)TrongtamgicvungBHE(H =90o)cHBE=50oHBE =90o-HBE=90o-50o=40o 0,5imHEM =HEB-MEB=40o-25o=15o 0,5imBMElgcngoitinhMca HEM NnBME=HEM +MHE =15o+90o=105o(nhlgcngoicatamgic) 0,5imBi 5: (4im)Trang 90Nm hc 2010-201120 0MABCD-Vhnha)Chngminh ADB= ADC(c.c.c) 1imsuyra DAB DAC 0,5imDo0 020 : 2 10 DAB 0,5imb) ABCcntiA,m020 A (gt)nn0 0 0(180 20 ) : 2 80 ABC ABCunn060 DBC 0,5imTiaBDnmgiahaitiaBAvBCsuyra0 0 080 60 20 ABD .TiaBMlphngiccagcABDnn010 ABM 0,5imXttamgicABMvBADc:ABcnhchung; 0 020 ; 10 BAM ABD ABM DAB Vy: ABM= BAD(g.c.g)suyraAM=BD,mBD=BC(gt)nnAM=BC 0,5im 4 Bi Ni dung cn t im1.1S hng th nht l (-1)1+1(3.1-1)1 S hng th hai l (-1)2+1(3.2-1) Dng tng qut ca s hng th n l: (-1)n+1(3n-1)1.2 A = (-3).17 = -51 1Trang 91Nm hc 2010-20112.123 4x y, 3y = 5z. Nu x-2y = 5 x= -15, y = -10, z = -60,5Nu x-2y = -5 x= 15, y = 10, z = 6 0,52.22 5x y 24 10x xy =9 x = 6 0,5Ta c 2x = 3z nn x1 = 6; y1 = 15; z1 = 4 v0,25x1 = -6; y1 = -15; z1 = -4 0,252.31 y zx+ +=2 x zy+ +=3 x yz+ =1x y z + +=2 0,5 x+y+z = 0,5 0, 5 1 0, 5 2 0, 5 3 x y zx y z + + = 2 0,5 x = 12; y = 56; z = - 560,53.13 8 9 1 2 9 1 22 3 4 9 1 1 2 9...... 1...a a a a a a a aa a a a a a a a+ + + + + + (v a1+a2++a9 0) 0,25 a1 = a2; a2 = a3; ;a9 = a1 0,25 a1 = a2 = a3== a93.2( ) ( )( ) ( )a b c a b c a b c a b ca b c a b c a b c a b c+ + + + + + + + =212bb (v b0) 0,25 a+b+c = a+b-c 2c = 0 c = 0 0,254.1t c1 = a1-b1; c2 = a2-b2;; c5 = a5-b50,25Xt tng c1 + c2 + c3 ++ c5 = (a1-b1)+( a2-b2)++( a5-b5) = 00,25 c1; c2; c3; c4; c5 phi c mt s chn0,25 c1. c2. c3. c4. c5 M2 0,254.2 AOE = BOF (c.g.c) O,E,F thng hng v OE = OF0,5 AOC = BOD (c.g.c) C,O,D thng hng v OC = OD EOD = FOC (c.g.c) ED = CF5Bi Ni dung cn t im1.1 S b chia = 4/11 0,5S chia = 1/11 0,25Kt qu = 4 0,251.2V 2x-272007 0 x v (3y+10)2008 0 y0,25 2x-272007 = 0 v (3y+10)2008= 00,25x = 27/2 v y = -10/3 0,51.3V 00ab 99 v a,b N 0,25Trang 92Nm hc 2010-2011200700 2007ab2007990,254472 110; 12>110; 13>110 19>110; 110 = 1100,51 1 1 1... 101 2 3 100+ + + + >0,53.2Ta c C = -18 - ( 2 6 3 9 x y + +) -180,5V 2 6 x 0;3 9 y + 00,25Max C = -18 2 6 03 9 0xy '+ x = 3 v y = -30,254.1 ABH = CAK (g.c.g) BH = AK4.2 MAH = MCK (c.g.c) MH = MK (1) gc AMH = gc CMK gc HMK = 900 (2)T (1) v (2) MHK vung cn ti Mp n s 6Cu1: Nhn tng v bt ng thc ta c : (abc)2=36abc+, Nu mt trong cc s a,b,c bng 0 th 2 s cn li cng bng 0+,Nuc 3s a,b,c khc 0 th chia 2 v cho abc ta c abc=36+, T abc =36 v ab=c ta c c2=36 nn c=6;c=-6+, T abc =36 v bc=4a ta c 4a2=36 nn a=3; a=-3Trang 93Nm hc 2010-2011+, T abc =36 v ab=9b ta c 9b2=36 nn b=2; b=-2-, Nu c = 6 th av b cng du nn a=3, b=2 hoc a=-3 , b=-2-, Nu c = -6 th av b tri du nn a=3 b=-2 hoc a=-3 b=2Tm li c 5 b s (a,b,c) tho mn bi ton(0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)Cu 2. (3)a.(1) ( 5x-3( -2 x>1*Nu 3x+1 x1 hoc x x 4 (0,25)(1)4-x+2x=3 => x=-1( tho mn k) (0,25)*4-x x>4 (0,25)(1) x-4+2x=3 x=7/3 (loi) (0,25)Cu3. (1) p dng ( a+b(( a( +( b( Ta cA=( x( +( 8-x( ( x+8-x( =8MinA =8 x(8-x) 0 (0,25)*' 0 80xx=>0 x 8 (0,25)*' 0 80xx=> '80xxkhng tho mn(0,25)Vy minA=8 khi 0 x 8(0,25)Cu4.Ta c S=(2.1)2+(2.2)2+...+ (2.10)2(0,5) =22.12+22.22+...+22.102=22(12+22+...+102) =22.385=1540(0,5)Cu5.(3)Trang 94AB MCDENm hc 2010-2011Chng minh: a (1,5)Gi E l trung im CD trong tam gic BCD c ME l ng trung bnh => ME//BD(0,25)Trong tam gic MAE c I l trung im ca cnh AM (gt) m ID//ME(gt)Nn D l trung im ca AE => AD=DE (1)(0,5)V E l trung im ca DC => DE=EC (2) (0,5)So snh (1)v (2) => AD=DE=EC=> AC= 3AD(0,25)b.(1)Trong tam gic MAE ,ID l ng trung bnh (theo a) => ID=1/2ME (1) (0,25)Trong tam gic BCD; ME l ng trung bnh => ME=1/2BD (2)(0,5)So snh (1) v (2) => ID =1/4 BD (0,25)----------------------------------------------------------------p n s 7Cu 1.Ta c. . .dadccbba (1) Ta li c.a c bc b adccbba+ ++ + (2)T (1) v(2) => dad c bc b a ,_+ ++ +3.Cu 2. A = a c bb a cc b a+++.= ( ) c b ac b a+ ++ +2.Nu a+b+c 0 => A = 21.Nua+b+c = 0=>A = -1.Cu 3.a). A = 1 + 25 x A Z thx- 2 l c ca 5.=> x 2 = (t1; t 5)*x = 3=>A = 6*x = 7=>A = 2*x = 1=>A = - 4*x = -3=>A = 0 b) A = 37+ x - 2 A Z thx+ 3 l c ca 7.Trang 95Nm hc 2010-2011=> x + 3 = (t1; t 7)*x = -2=>A = 5*x = 4=>A = -1*x = -4 =>A = - 9*x = -10=>A = -3 .Cu 4.a). x =8hoc - 2 b). x =7hoc - 11c). x =2.Cu 5. ( T v hnh) MHK l cnti M .Tht vy: ACK = BAH. (gcg)=>AK = BH . AMK = BMH (g.c.g) =>MK = MH.Vy: MHK cnti M .--------------------------------------------------------------------p n s 8Cu 1:Gi x, y, z l di 3 cnh tng ngvi cc ng cao bng 4, 12, a. Ta c: 4x= 12y = az= 2Sx= S/2 ; y = S/6;z = 2S/a(0,5 im)Do x-y < z< x+y nn32 2626 226 2< < + < < aS SaS S S(0,5 im) 3, a , 6 Do a Nnn a=4 hoc a= 5. (0,5 im)2. a. Tdcba d c cb a ad cb acad cb adbca (0,75 im)b. dcbadd cb b ad cb adbd cb adbca ++++ ++ (0,75 im)Cu 2: V tch ca 4 s :x2 1 ; x2 4; x2 7;x2 10 l s m nn phi c1 s m hoc 3 s m.Ta c :x2 10< x2 7< x2 4< x2 1. Xt 2 trng hp:+ C 1 s m:x2 10 1 2 32 3 4x y z v 2x + 3y - z = 50(0,5)=> x = 11, y = 17, z = 23. (0,5)Cu 3(2): Cc phn s phi tm l: a, b, c ta c : a + b + c = 21370v a : b : c = 345: : 6: 40: 255 1 2 (1) => 9 12 15, ,35 7 14a b c (1)Cu 4(3):K DF // AC ( F thuc BC ) (0,5 )=> DF = BD = CE (0,5 ) =>IDF = IFC ( c.g.c ) (1 )=> gc DIF = gc EIC => F, I, C thng hng => B, I, C thng hng(1)Cu 5(1):=> 7.2 1 1(14 1) 77xy xy+ + => (x ; y ) cn tm l ( 0 ; 7 )----------------------------------------------------------------------p n s 10Cu 1: a)Ta c: 21112 . 11 ; 31213 . 21 ;41314 . 31 ; ;1001991100 . 99 1 Vy A = 1+10099100111001991991....31312121 ,_++ + ,_++ ,_+b) A = 1+ ,_+ + ,_+ ,_+ ,_221 . 20201....25 . 44124 . 33123 . 221=Trang 98Nm hc 2010-2011= 1+ ( ) + + + + + + + 21 ... 4 3 221221...2423= ,_1222 . 2121= 115.Cu 2: a) Ta c:4 17 > ; 5 26 >nn 1 5 4 1 26 17 + + > + + hay 10 1 26 17 > + +Cn99 < 10 .Do : 99 1 26 17 > + +b) ;10111>10121>;10131>; ..; 1011001.Vy: 10101. 1001001....312111 > + + + +Cu 3: Gi a,b,ca l cc ch s ca s c ba ch scn tm . V mi ch s a,b,ca khng vt qu 9 v ba ch s a,b,ca khng th ng thi bng 0 , v khi ta khng c s c ba ch s nn: 1 a+b+c27Mt khc s phi tm l bi ca 18 nn a+b+c =9 hoc a+b+c = 18 hoc a+b+c=17Theo gi thit, ta c:6 3 2 1c b a c b a + + Do : ( a+b+c) chia ht cho 6Nn : a+b+c =18 36183 2 1 c b a a=3; b=6 ; ca =9V s phi tm chia ht cho 18 nnch s hng n v ca n phi l s chn.Vy cc s phi tm l: 396; 936.Cu 4:a) V AH BC; ( H BC) ca ABC+ hai tam gic vung AHB v BIDc:BD= AB (gt)Gc A1= gc B1( cng ph vi gcB2) AHB= BID ( cnh huyn, gc nhn)AH BI (1) v DI= BH+ Xt hai tam gic vung AHC v CKE c: Gc A2= gc C1( cng ph vi gc C2)AC=CE(gt)Trang 99Nm hc 2010-2011 AHC= CKB ( cnh huyn, gc nhn) AH= CK (2)t (1) v (2) BI= CK v EK = HC.b) Ta c: DI=BH ( Chng minh trn)tng t: EK = HCT BC= BH +Hc= DI + EK.Cu 5: Ta c:A = 1 2001 + x x=2000 1 2001 1 2001 + + x x x xVy biu thc cho t gi tr nh nht l 2000 khi x-2001 v 1-x cng du, tc l :1x 2001biu im :Cu 1: 2 im . a. 1 imb. 1imCu 2: 2 im : a. 1 im b . 1 im .Cu 3 : 1,5 imCu 4: 3 im :a. 2 im ; b. 1 im .Cu 5 : 1,5 im .---------------------------------------------------------------------p n s11Cu1:a,(1)0 4534913245132541326313272 ++ +++ +++ +++ ++x x x x x (0,5 )...... 0 )513241325132613271)( 329 ( + + + + + x329 0 329 + x x (0,5 )b, a.Tm x,bit: |5x - 3| - x = 7 5 3 7 x x + (1)(0,25 )K:x -7(0,25 )( )( )5 3 715 3 7x xx x + +.(0,25 )Vy c hai gi tr x tha mn iu kin u bi. x1 = 5/2;x2= - 2/3 (0,25).Cu 2:Trang 100Nm hc 2010-2011a, 2007 4 3 271.....717171711 + + + S ; 2006 3 271.....7171711 7 7 + + S(0.5) 2007717 8 S87172007 S(0,5)b,! 100 1 100.......! 3 1 3! 2 1 2! 10099......! 43! 32! 21 + ++ + + + +(0,5)................... 1! 10011 < (0,5)c, Ta c +23n) 2 2 ( 3 3 2 3 22 2 2 n n n n n n n + ++ + +(0,5)................. ( ) 10 2 3 10 10 . 2 10 . 3 5 . 2 10 . 32 2 n n n n n n(0,5)Cu 3: Gi di 3 cnh l a , b, c, 3 chiu cao tng ng l x, y, z, din tch S ( 0,5 )xSa2ySb2 zSc2 (0,5) zSySxS c b a4232224 3 2 (0,5)3 4 64 3 2z y xz y x vy x, y, z t l vi6 ; 4 ; 3(0,5)Cu4: GT; KL; Hnh v(0,5)a,Gc AIC = 1200 (1 )b,LyAC H : AH = AQ .............. IP IH IQ (1 )Cu5:B ; LN ( ) 3 1 2 ;2+ n LN BNNV( ) ( ) 3 3 1 2 0 12 2 + n n t NN khi bng 3 (0,5)Du bng xy ra khi 1 0 1 n nvy B ; LN 31 B v 1 n(0,5)-------------------------------------------------------------p n s 12Cu 1 : 3 im . Mi cu 1 imd) (x-1)5 = (-3)5 x-1 = -3 x = -3+1 x = -2e) (x+2)(151141131121111 + +) = 0151141131121111 + + 0 x+2 = 0 x = 2f) x - 2 x= 0 ( x )2- 2 x= 0 x ( x - 2) = 0 x= 0 x = 0hocx - 2 = 0 x= 2 x = 4Trang 101Nm hc 2010-2011Cu 2 : 3 im . Mi cu 1,5 ima) 8145 + yx,8182 5 +yx ,82 1 5 yxx(1 - 2y) = 40 1-2y l c l ca 40 . c l ca 40 l :t1 ;t5 .p s :x = 40 ; y = 0x = -40 ; y = 1x = 8 ; y = -2x = -8 ; y = 3b) Tm xz AZ. A= 34131+ +x xxA nguyn khi 34 x nguyn 3 x(4) = {-4 ; -2 ;-1; 1; 2; 4}Cc gi tr ca x l : 1 ; 4; 16 ; 25 ; 49 .Cu 3 : 1 im23 5 x - 2x = 143 5 x=x + 7 (1)K:x -7(0,25 )( )( )5 3 715 3 7x xx x + +.(0,25 )Vy c hai gi tr x tha mn iu kin u bi. x1 = 5/2;x2= - 2/3 (0,25). Cu4.(1.5 im)Cc gc A, B , C t l vi 7, 5, 3121518015 3 5 70 + + C B A C B AA= 840 gc ngoi ti nh A l 960B =600 gc ngoi ti nh B l 1200C = 360 gc ngoi ti nh C l 1440 Cc gc ngoi tng ng t l vi 4 ; 5 ; 6b)1) AE = AD ADE cn 1E DE EDA 1E = 01802A (1) ABC cn B C 1AB C= 01802A (2)T (1) v (2) 1E ABC Trang 102Nm hc 2010-2011ED // BCb) XtEBC vDCB c BC chung (3) EBC DCB (4)BE = CD (5)T (3), (4), (5) EBC =DCB (c.g.c) BEC CDB = 900 CE AB ..p n s 13Bi 1: 3 ima, Tnh:A = 11160.36471300475.11121 .3311 1160).41915(100175310(1112)71767183(331 = 1815284284551001.332841001553357 34110011001100110561119331 b, 1,5 im Ta c:+) 1 + 4 +7 ++ 100= ( 1+100) + ( 4 + 97) +.+ ( 49+ 52)= 101 . 34 = 143434 cp+) 1434 410 = 1024+) ( 18 . 123 + 9 . 436 . 2 + 3 . 5310. 6 )= 18 . ( 123 + 436 + 5310 )= 18 . 5869 =105642Vy A = 105642 : 1024103,17Bi 2: 2 imGii s cn tm l x, y, z. S nh l x , s ln nht l z. Ta c: xy z (1)Theo gi thit:21 1 1 + +z y x(2). Do (1) nn z =x z y x3 1 1 1 + +Trang 103Nm hc 2010-2011Vy: x = 1. Thay vo (2) , c: y z y211 1 +Vy y = 2. T z = 2.Ba s cn tm l 1; 2; 2.Bi 3:2 imC 9 trang c 1 ch s. S trang c 2 ch s l t 10 n 99 nn c tt c 90 trang. Trang c 3 ch s ca cun sch l t 100 n 234, c tt c 135 trang. Suy ra s cc ch s trong tt c cc trang l:9 + 2 . 90 + 3. 135 = 9 + 180 + 405 = 594Bi 4 : 3 imTrn tia EC ly im D sao cho ED = EA.Hai tam gic vungABE =DBE ( EA = ED, BE chung)Suy ra BD = BA ; BAD BDA .Theo gi thit: EC EA = A BVy EC ED = AB Hay CD = AB (2)T (1) v (2) Suy ra: DC = BD.V tia ID l phn gic ca gc CBD ( I BC ).Hai tam gic:CID vBID c :ID l cnhchung,CD = BD ( Chng minh trn). CID=IDB ( v DI l phn gic ca gc CDB )VyCID = BID ( c . g . c) C =IBD. Gi C l BDA=C+IBD= 2C = 2 ( gc ngoi ca BCD) m A=D( Chng minh trn) nn A= 2 + 2= 900 = 300 .Do ; C=300 v A = 600----------------------------------------------H ng dn gii s14 Bi 1.a.Xt 2 trng hp :*5 x ta c : A=7.* 5 x > hay A > 7.Vy : Amin = 7 khi5 x .Bi 2. a.t : A =2 2 2 21 1 1 1.......5 6 7 100+ + + +Trang 104Nm hc 2010-2011Ta c :*A < 1 1 1 1.........4.5 5.6 6.7 99.100+ + + + = 1 1 1 1 1 1.....4 5 5 6 99 100 + + + = 1 1 14 100 4 1 1 1 1 1 1 1.........5.6 6.7 99.100 100.101 5 101 6+ + + + >.b.Ta c : 2 9 5 17 33 3 3a a aa a a+ ++ + + += 4 263aa++ == 4 12 14 4( 3) 14 1443 3 3a aa a a+ + + + ++ + + l s nguynKhi (a + 3) l c ca 14 m (14) = 1;2;7;14 t t t t.Ta c : a = -2;- 4;- 1; - 5; 4 ; - 10; 11 ; -17.Bi 3. Bin i :( ) 12 1 30. A n nn + + ( ) 6 1 30 6 A n nn n1 + ]M M* ( ) 1 30 nn n n M M n (30) hay n {1, 2 , 3, 5 , 6 , 10 , 15 , 30}.* ( ) ( ) 30 6 16 1 3 nn nn M M M+ { } 3 3, 6,15, 30 . n n M+( ) { } 1 3 1,10 . n n M n {1 , 3 , 6 , 10 , 15 , 30}.-Th tng trng hp ta c : n = 1, 3, 10, 30 tho mn bi ton.Bi 4.-Trn Oy ly M sao cho OM = m. Ta c :N nm gia O, M v MN = OM.-Dng d l trung trc ca OM v Oz lphn gic ca gc xOy chng ct nhau ti D.-' ( . . ) ODM MDNc g c MD ND V VD thuc trung trc ca MN.-R rng : D c nh. Vy ng trung trc ca MN i qua D c nh.Bi 5. -Dng tng qut ca a thc bc hai l :( )2f x ax bx c + +(a0).- Ta c :( ) ( ) ( )21 1 1 f x a x b x c + + .- ( ) ( ) 1 2 f x f x ax a b x + 2 10ab a ' 1212ab 'Vy a thc cn tm l :( )21 12 2f x x x c + + (c l hng s).Trang 105 xz ddmn i ym'oNm hc 2010-2011p dng :+ Vi x = 1 ta c :( ) ( ) 1 1 0 . f f + Vi x = 2 ta c :( ) ( ) 1 2 1 . f f .+ Vi x = n ta c :( ) ( ) 1 . n f n f n S = 1+2+3++n =( ) ( ) 0 f n f = ( )212 2 2nnn nc c++ + .L u :Hc sinh gii cch khc ng vn cho im ti a. Bi hnh khng v hnh khng chm im.--------------------------------------------------------------------p n s 15Cu1 (lm ng c 2 im)Ta c: 228 20x xx x+ =222 10 20x xx x x + = 2( 2)( 10)x xx x + (0,25)iu kin(x-2)(x+10) 0 x 2;x -10 (0,5)Mt khc 2 x = x-2nu x>2-x + 2 nux< 2(0,25)* Nu x> 2 th 2( 2)( 10)x xx x +=( 2)( 2)( 10)xxx x + = 10xx +(0,5)* Nu x 0; y >0 ; z >0)Trang 106Nm hc 2010-2011Theo ra ta c{94(1)3 4 5 (2)x y zx y z++ (0,5)BCNN (3,4,5) = 60T (2) 360x=460y=560zhay 20x=15y=12z(0,5)p dng tnh cht dy t s bng nhau ta c :20x=15y=12z= 20 15 12x y z + ++ += 9447=2(0,5)x= 40, y=30 v z =24 (0,5)S hc sinh i trng cy ca 3 lp 7A, 7B, 7C ln lt l 40, 30, 24.Cu 3 (lm ng cho 1,5) 200610 539+ l s t nhin 102006 + 53 M9(0,5) 102006 + 53 M9102006 + 53c tng cc ch s chia ht cho 9m 102006 + 53= 1+ 0 +0 +.........+0 + 5+3= 9M9 102006 + 53 M9hay200610 539+ l s t nhin(1)Cu 4 (3)- V c hnh, ghi GT, KL c 0,25a, ABC c 1 2A A (Az l tia phn gic caA ) 1 1A C (Ay // BC, so le trong) 2 1A C ABC V cn ti Bm BK AC BK l ng cao ca cn ABC BK cng l trung tuyn ca cn ABC (0,75)hay K l trung im ca ACb, Xt ca cn ABH vvung BAK.C AB l cng huyn(cnh chung) 02 1( 30 ) A B V {020 0 0130290 60 30AAB vung ABH = vung BAK BH = AK m AK = 2 2AC ACBH (1)Trang 107Nm hc 2010-2011c, AMC vung ti M c AK = KC = AC/2(1) MK l trung tuyn thuc cnh huyn KM = AC/2(2)T (10 v (2) KM = KC KMC cn.Mt khc AMC c 0 0 0 0 090 A=30 90 30 60 M MKC AMC u (1)Cu 5. Lm ng cu 5 c 1,5Xy dng s cy v gii bi tonp n : Ty t gii nht, Nam gii nh, ng gii 3, Bc gii 4-------------------------------------p n s 16Cu 1: (2)a) Xt khong 32 xc x = 4,5 ph hp 0,25 Xt khong 32< xc x = - 45ph hp 0,25 b) Xt khong 23 x c x > 4 0,2Xt khong 23< x c x < -1 0,2Vy x > 4 hoc x < -1 0,1c) Xt khong 31 x Ta c 3x - 1 738 x Ta c 3831 xXt khong 31< x Ta c -3x + 17 2 xTa c 312 xVy gi tr ca x tho mn bi l 382 xCu 2:a) S = 1+25 + 252 +...+ 25100 0,3Trang 108Nm hc 2010-20111 25 25 2425 ... 25 25 25101101 2 + + + S S SS 0,3Vy S = 241 25101 0,1b) 430= 230.230 = (23)10.(22)15 >810.315> (810.310)3 = 2410.3 0,8Vy 230+330+430> 3.224 0,2Cu 3:a) Hnh a.AB//EF v c hai gc trong cng pha b nhauEF//CD v c hai gc trong cng pha b nhauVy AB//CDb) Hnh b.AB//EF V c cp gc so le trong bng nhau 0,4CD//EF v c cp gc trong cng pha b nhau 0,4Vy AB//CD 0,2Cu 4:(3)a)MN//BC MD//BD D trung im AP 0,3 BP va l phn gic va l trung tuyn nn cng l ng cao BDAP 0,2Tng t ta chng minh c BE AQ 0,5 b) AD = DPBDE DBP (g.c.g) DP = BE BE = AD 0,5 MD ME c g c MAD MBE ) . . ( 0,3BP = 2MD = 2ME = BQTrang 109Nm hc 2010-2011Vy B l trung im ca PQ 0,2c)BDE vung B, BM l trung tuyn nn BM = ME 0,4ADB vung D c DM l trung tuyn nn DM = MA 0,4DE = DM + ME = MA + MB 0,2Cu 5: 1A = x +4101A ln nht x 410ln nht0,3Xt x > 4 thx 410 < 0Xt 4 < x th x 410> 0a ln nht 4 - x nh nht x = 3 0,6------------------------------------------------------------------------------p n s 17Cu 1: ( mi 0,5 im ).a/.4 3 x +- x = 15. b/. 3 2 x - x> 1.4 3 x + = x + 15 3 2 x > x + 1* Trng hp 1: x -34 , ta c: * Trng hp 1: x 23, ta c:4x + 3 = x + 153x - 2 > x + 1 x = 4 ( TMK). x > 32 ( TMK).* Trng hp 2: x < - 34 , ta c: * Trng hp 2: x < 23, ta c:4x + 3 = - ( x + 15) 3x 2 < - ( x + 1) x = - 185 ( TMK). x < 14 ( TMK)Trang 110Nm hc 2010-2011Vy: x = 4 hoc x = - 185. Vy: x > 32 hoc x < 14.c/. 2 3 x + 5 5 2 3 5 x + 4 1 x Cu 2:a/.Ta c: A= (- 7) + (-7)2 + + (- 7)2006 + (- 7)2007 ( 1 ) (- 7)A = (-7)2 + (- 7)3 + + (- 7)2007 + (- 7)2008 ( 2)8A =(- 7) (-7)2008Suy ra:A = 18.[(- 7) (-7)2008 ] = - 18( 72008+ 7 )* Chng minh: A M43.Ta c: A= (- 7) + (-7)2 + + (- 7)2006 + (- 7)2007 , c 2007 s hng. Nhm 3 s lin tip thnh mt nhm (c 669 nhm), ta c:A=[(- 7) + (-7)2 + (- 7)3] + + [(- 7)2005 + (- 7)2006 + (- 7)2007]= (- 7)[1 + (- 7) + (- 7)2] + + (- 7)2005. [1 + (- 7) + (- 7)2]= (- 7). 43 + + (- 7)2005. 43= 43.[(- 7) + + (- 7)2005] M 43Vy : A M 43b/. * iu kin :Nu m M 3 v n M 3 th m2 M 3, mn M 3 v n2 M 3, do : m2+ mn + n2 M 9.* iu kin cn:Ta c: m2+ mn + n2 = ( m - n)2 + 3mn.(*)Nu m2+ mn + n2 M 9 thm2+ mn + n2 M 3, khi t (*),suy ra: ( m - n)2 M 3 ,do ( m - n) M 3v th( m - n)2 M 9 v 3mn M 9 nn mn M 3 ,do mt trong hai s m hoc n chia ht cho 3 m ( m - n) M 3nn c 2 s m,n u chia ht cho 3.Cu 3:Gi di cc cnh tam gic l a, b, c ; cc ng cao tng ng vi cc cnh l ha , hb , hc .Ta c: (ha +hb) : ( hb + hc ) : ( ha + hc ) = 3 : 4 : 5Hay: 13(ha +hb) = 14( hb + hc ) =15( ha + hc ) = k ,( vi k 0).Trang 111ABCDNm hc 2010-2011Suy ra: (ha +hb) = 3k ; ( hb + hc ) = 4k ; ( ha + hc ) = 5k .Cng cc biu thc trn, ta c: ha + hb + hc = 6k.T ta c:ha = 2k;hb =k;hc = 3k.Mt khc, gi S l din tchABC V, ta c:a.ha = b.hb =c.hc a.2k = b.k = c.3k3a = 6b = 2cCu 4:Gi s DC khng ln hn DB hay DC DB.* Nu DC = DB thBDC Vcn ti D nn DBC = BCD.Suy ra:ABD = ACD.Khi ta c: ADB V=ADC V(c_g_c) . Do : ADB = ADC ( tri vi gi thit).* Nu DC < DB th trongBDC V , ta c DBC < BCD m ABC = ACB suy ra:ABD >ACD ( 1 ) .XtADB VvACD Vc: AB = AC ; AD chung ; DC < DB.Suy ra: DAC < DAB ( 2 ).T (1) v (2) trongADB VvACD Vta li c ADB < ADC , iu ny tri vi gi thit.Vy:DC > DB.Cu 5: ( 1 im)p dng bt ng thc: x y x- y, ta c:A = 1004 x -1003 x + ( 1004) ( 1003) x x + = 2007Vy GTLN ca A l: 2007.Du = xy ra khi: x -1003.-----------------------------------------------------------------H ng dn chm 18 Trang 112Nm hc 2010-2011Cu 1-a (1 im ) Xt 2 trng hp 3x-2 0. 3x -2 kt lun : Khng c gi tr no ca x tho mn.b-(1 im ) Xt 2 trng hp 2x +5 0 v 2x+5 kt lun.Cu 2-a(2 im ) Gi s cn tm l abcabc18=> abc9. Vy (a+b+c)9(1)Ta c : 1 a+b+c27(2)T (1) v (2) suy ra a+b+c =9 hoc 18 hoc 27 (3)Theo bi ra 1a= 2b=3c = 6c b a + + (4)T (3) v (4) => a+b+c=18.v t (4) => a, b, c mabc2 => s cn tm : 396, 936.b-(1 im )A=(7 +72+73+74) + (75+76+77+78) + ...+ (74n-3+ 74n-2+74n-1+74n).= (7 +72+73+74) . (1+74+78+...+74n-4).Trong : 7 +72+73+74=7.400 chia ht cho 400 . Nn A400Cu 3-a (1 im ) T C k Cz//By c : 2C +CBy=2v (gc trong cng pha) (1) 1C +CAx=2v V theo gi thit C1+C2 + + = 4v =3600.Vy Cz//Ax.(2)T (1) v (2) => Ax//By.Cu 4-(3 im)ABC cn, ACB =1000=> CAB = CBA =400.Trn AB ly AE =AD. Cn chng minh AE+DC=AB (hoc EB=DC)AED cn, DAE = 400: 2 =200.=> ADE =AED = 800 =400+EDB (gc ngoi caEDB)=> EDB =400 => EB=ED (1)Trn AB ly C sao cho AC = AC. C CAD = CAD ( c.g.c)D ACD = 1000 vDCE = 800.VyDCE cn => DC =ED (2)T (1) v (2) c EB=DC.A CEBM DC =DC. Vy AD +DC =AB.Cu 5 (1 im).S=(-3)0+(-3)1 + (-3)2+(-3)3+...+ (-3)2004.Trang 113Nm hc 2010-2011-3S= (-3).[(-3)0+(-3)1+(-3)2 + ....+(-3)2004]= (-3)1+ (-3)2+ ....+(-3)2005]-3S-S=[(-3)1 + (-3)2+...+(-3)2005]-(3)0-(-3)1-...-(-3)2005.-4S = (-3)2005 -1. S = 41 ) 3 (2005 =41 32005+---------------------------------------------------------p n 19Bi 1: Ta c : - 2161121201301421561721901 = - (10 . 919 . 818 . 717 . 616 . 515 .. 414 . 313 .. 212 . 11+ + + + + + + +) 1= - (101919181.....413131212111 + + + + + ) 1=- (10111)= 109 0,5Bi 2: A = x x + 5 2Vi x3 0,5Vi 2x5 th A = x-2 x+5 = 3 0,5Vi x>5 th A = x-2 +x 5 = 2x 7 >30,5So snh cc gi tr ca A trong cckhong ta thy gi tr nh nht ca A = 3 2x5 1Bi 3: a. Trn tia i ca tia OC ly im N saocho ON = OC .Gi M l trung im ca BC.nn OMl ng trung bnh catam gic BNC.Do OM //BN, OM= 21 BNDo OM vung gc BC => NB vung gc BCM AH vung gcvi BC v th NB // AH (1)Tng t AN//BHDo NB = AH. Suy ra AH = 2OM (1)Trang 114AC BOGHNm hc 2010-2011b. Gi I, K theo th t l trung im ca AG v HG thIK l ng trung bnh ca tam gic AGH nn IK// AHIK = 21 AH => IK // OMv IK = OM ; KIG =OMG (so le trong)IGK= MGO nn GK = OG v IGK =MGOBa im H,G, O thng hng1Do GK = OG m GK= 21 HG nn HG = 2GOng thng qua 3 im H, G, O c gi l ng thng le.1Bi 4: Tng cc h s ca mt a thc P(x) bt k bng gi tr ca a thc ti x=1. Vy tng cc h s ca a thc:0,5P(x) = (3-4x+x2)2006 . (3+4x + x2)2007Bng P(1) = (3-4+1)2006 (3+4+1)2007= 00,5------------------------------------------------------------p n 20Cu 1: Ta c:220 0 (mod2) nn 22011969 0 (mod2)119 1(mod2) nn 11969220 1(mod2)69 -1 (mod2) nn 69220119 -1 (mod2)Vy A 0(mod2) hay A M2 (1)Tng t:AM 3 (1)A M17(1)V 2, 3, 17 l cc s nguyn t A M 2.3.17 = 102Cu 2: Tm xa) (1,5)Vi x < -2 x = -5/2 (0,5)Vi -2 x 0 khng c gi tr x no tho mn(0,5)Vi x > 0x = (0,5)Trang 115Nm hc 2010-2011b) (1,5) Vi x < -2 Khng c gi tr x no tho mn(0,5)Vi -2 x 5/3 Khng c gi tr x no tho mn(0,5)Vi x > 5/3 x = 3,5(0,5)Bi 3:a) D dng chng minh c IH = 0MAIH // 0M do 0MN = HIK (g.c.g) IEDo : IHQ = M0Q (g.c.g) QH = Q0FH NQI = QM Pb) DIM vung c DQ l ng trungK QOtuyn ng vi cnh huyn nnRQD = QI = QMB D MCNhng QI l ng trung bnh ca 0HA nn c) Tng t: QK = QN = QE = OB/2QR = QP = QF = OC/2Bi 4(1):V 3|x-5| 0 x RDo A = 10 - 3|x-5| 10Vy A c gi tr ln nht l 10 |x-5| = 0 x = 5----------------------------------------------------------------p n 21Bi 1.iu kin x 0(0,25)a) A = - 79(0,5)b) 3 + x > 0 A = -1 3 5 x x x = 1 (0,5)c) Ta c: A = 1 - 38+ x.(0,25) A Z th 3 + x l c ca 8 x = {1; 25} khi A = {- 1; 0} (0,5)Bi 2.Trang 116Nm hc 2010-2011a) Ta c: 1 7 x x 32 ; 31) 1 ( 70 12 ' ' xx xxx xx(1)b) Ta c: 2M = 2 22 + 23 24 + - 22006 + 22007 (0,25) 3M = 1 + 22007(0,25) M = 31 22007+(0,5)c) Ta c: A = x4 + 2x2 +1 1 vi mi x PCM.(1)Bi 3. Ta c: 00 180301 2 3 6A B C 0 0 0 30 ; 60 ; 90 A B C (0,5)Vy tam gic ABC l tam gic vung ti C (0,5)Bi 4.GT, KL (0,5)a) Gc AIC = 1200 (1)b) Ly H AC sao cho AH = AN (0,5)T chng minh IH = IN = IM(1)Bi 5.A = 1 + x 62000 (0,5)AMax6 x > 0 v nh nht 6 x = 1 x = 5.Vy x = 5 tho mn iu kin bi ton khi A Max= 2001 (0,5)--------------------------------------------------------------------p n 22Cu 1: (2.5)a. a1.55 40 15 20 152121.2141.21,_ ,_,_ ,_,_(0.5)a2.30 2591,_,_31: = 30 5031,_,_31: = 203,_(0.5)Trang 117Nm hc 2010-2011b. A = 31) 5 1 ( 3 . 2) 3 1 .( 3 . 220 . 6 3 . 26 . 2 9 . 48 108 108 8 109 4 5++(0.5)c. c1. 337= 0.(21) c2.227= 0,3(18) (0.5)c3. 0,(21) = 3379921 ; c4.5,1(6) = 561(0.5)Cu 2: (2)Gi khi lng ca 3 khi 7, 8, 9 ln lt l a, b, c (m3)a + b + c = 912 m3. (0.5)S hc sinh ca 3 khi l : 2 , 1a ;4 , 1b ;6 , 1cTheo ra ta c: 2 , 1 1 , 4 . 3a b v 6 , 1 . 5 4 , 1 . 4c b(0.5)206 , 1 . 15 4 , 1 . 12 2 , 1 . 4 c b a(0.5)Vy a = 96 m3 ;b = 336 m3 ;c = 480 m3.Nn s HS cc khi 7, 8, 9 ln lt l: 80 hs, 240 hs, 300 hs. (0.5)Cu 3: ( 1.5):a.Tm max A.Ta c: (x + 2)2 0 (x = 2)2 + 4 4 Amax= 43 khi x = -2 (0.75)b.Tm min B.Do (x 1)2 0 ; (y + 3)20 B1Vy Bmin= 1 khi x = 1 v y = -3 (0.75)Cu 4: (2.5)K CHct MB ti E. Ta c EAB cn ti E EAB =300 EAM = 200 CEA = MAE = 200 (0.5)Do ACB = 800 ACE = 400 AEC = 1200 ( 1 )(0.5)Mt khc: EBC = 200 v EBC = 400 CEB = 1200( 2 ) (0.5)T ( 1 ) v ( 2 ) AEM = 1200Do EAC = EAM (g.c.g) AC = AM MAC cn ti A (0.5)V CAM = 400 AMC = 700. (0.5)Cu 5: (1.5)Gi s a2 v a + b khng nguyn t cng nhau a2 v a + bTrang 118E300100MCB A HNm hc 2010-2011Cng chia ht cho s nguyn t d: a2 chia ht cho d a chia htcho d v a + b chia ht cho d b chia hta cho d (0.5)(a,b) = d tri vi gi thit.Vy (a2,a + b) =1. (0.5)------------------------------------------------------- 23Cu I :1) Xc nh a, b ,c654 32 1 + c b a=224 12 1020 9 5 4 3 524) 5 ( 412) 3 ( 310) 1 ( 5 + + c b a c b a=> a = -3 ; b = -11; c = -7.Cch 2 :654 32 1 + c b a = t ; sau rt a, b ,c thay vo tm t =- 2 tm a,b,c.2) Chng minht dcba = k => a= kb ; c = kdThay vo cc biu thc :03 25 33 25 33 25 3 23 25 3 22 222 222 2++ ++ ++ ++ kk kkk kcd dd cd cab bb ab a=> pcm.Cu II: Tnh:1) Ta c :2A= 2(99 . 971....7 . 515 . 31+ + +) = 993299131991971.....71515131 + + + =>A =99162) B = = 51 50 3 23131.....313131 + + + = ) 3 ( 1) 3 ( 1.....) 3 ( 1) 3 ( 1) 3 ( 151 50 3 2++ +++) 3 ( 1) 3 ( 1.....) 3 ( 1) 3 ( 1) 3 ( 152 51 4 3 2++++ => B31 ) 3 ( 13152= 525131 3 => B = 51513 . 4) 1 3 ( Cu IIITrang 119Nm hc 2010-2011Ta c : 0.2(3) = 0.2 + 0.0(3) = +102.1010,(1).3 =91.103102+ =3070,120(32) = 0,12 + 0,000(32) =0,12+10001.0,(32)= 0,12+10001.0,(01).32 =991.10003210012+=123751489Cu IV :Gi a thc bc hai l : P(x) = ax(x-1)(x-2) + bx(x-1)+c(x-3) + dP(0) = 10 => -3c+d =10 (1)P(1) = 12 => -2c+d =12 =>d =12+2c thay vo (1) ta c -3c+12+2c =10 =>c=2 , d =16P(2)= 4 => 2b -2+16 = 4 > b= -5P(3) = 1 => 6a-30 +16 =1 => a = 25Vy a thc cn tm l : P(x) =16 3 2 1 5 2 125+ + ) ( ) ( ) )( ( x x x x x x=> P(x) = 325x-10 122252+ + x xCu V:a) D thyADC =ABE ( c-g-c) => DC =BE .V AE AC; AD ABmt khc gc ADC = gc ABE=> DC Vi BE.b) Ta c MN // DC v MP // BE => MN MPMN = 21DC = 21BE =MP;Vy MNP vung cn ti M.---------------------------------------------------------p n 24Bi 1:a)A = 3 3 3 3 3 3 38 10 11 12 2 3 45 5 5 5 5 5 58 10 11 12 2 3 4 + + + + + + (0,25) Trang 120Nm hc 2010-2011A =1 1 1 1 1 1 13 38 10 11 12 2 3 41 1 1 1 1 1 15 58 10 11 12 2 3 4 _ _ + + + , ,+ _ _ + + + , , (0,25)A = 35 + 35 = 0 (0,25)b) 4B = 22 + 24 + ... + 2102(0,25) 3B = 2102 1;B = 1022 13 (0,25)Bi 2:a) Ta c 430 = 230.415(0,25)3.2410 = 230.311(0,25)m 415 > 311 430 > 311 230 + 330 + 430 > 3.2410(0,25)b) 4 = 36 >2933 >14 (0,25)36 +33 >29 +14 (0,25)Bi 3:Gi x1, x2 x3 ln lt l s ngy lm vic ca 3 my 1 2 33 4 5x x x (1) (0,25)Gi y1, y2, y3 ln lt l s gi lm vic ca cc my 1 2 36 7 8y y y (2) (0,25)Gi z1, z2, z3 ln lt l cng sut ca 3 my 5z1 = 4z2 = 3z3 1 2 31 1 15 4 3z z z (3) (0,25)Mx1y1z1 + x2y2z2 + x3y3z3 = 359 (3) (0,25)T (1) (2) (3) 1 1 1 2 2 2 3 3 33951518 40 39575 3 15x y z x y z x y z (0,5) x1y1z1 = 54; x2y2z2 = 105; x3y3z3 = 200 (0,25)Vy s thc mi i ln lt l 54, 105, 200 (0,25)Bi 4:a)EAB = CAD (c.g.c) (0,5)Trang 121Nm hc 2010-2011 ABM ADM (1) (0,25)Ta c + BMC MBD BDM (gc ngoi tam gic) (0,25) 0 0 060 60 120 BMC MBA BDM ADM BDM + + + + (0,25)b) Trn DM ly F sao cho MF = MB (0,5)FBM u (0,25)DFB AMB(c.g.c) (0,25) 0120 DFB AMB (0,5)Bi 6: Ta c12 (2) 3. ( ) 42x f f + (0,25)1 1 1( ) 3. (2)2 2 4x f f + (0,25) 47(2)32f (0,5)------------------------------------------------------- p n 25 Cu 1a.Nu x0 suy ra x = 1 (tho mn)Nu < 0 suy ra x = -3(tho mn)b. ' 6 316 32161xyx xy; hoc' 6 31xy;hoc23 3yx ' hoc33 2yx ' ;hoc63 1yx ' ; hoc63 1yx ' hoc23 3yx ' ;hoc33 2yx ' T ta c cc cp s (x,y) l (9,1); (-3, -1) ; (6, 2) ; (0,- 2) ; (5, 3) ; (1, -3) ; (4, 6); (2, -6)c. T 2x = 3y v 5x = 7z bin i v3 7 5 3 7 5 30221 14 10 61 89 50 63 89 50 15x y z x y z x y z + + x = 42; y = 28; z = 20Trang 122MABCDEFNm hc 2010-2011Cu 2c. A l tch ca 99 s m do 2 2 2 2 21 1 1 1 1.3 2.4 5.3 99.1011 1 1 .... 14 9 16 100 2 3 4 1001.2.3.2....98.99 3.4.5...99.100.101 101 1 12.3.4...99.100 2.3.4......99.100 200 2 2AA _ _ _ _ , , , , > < g g ggggd. B = 1 3 4 413 3 3x xx x x+ + + B nguyn( ) 44 33nguen xx U{ } 4; 25;16;1; 49 x Cu 3Thi gian i thc t nhiu hn thi gian d nhGi vn tc i d nh t C n B l v1 == 4km/hVn tc thc t i t C n B l V2 = 3km/hTa c: 1 1 12 2 24 33 4V t VvaV t V (t1 l thi gian i AB vi V1; t2 l thi gian i CB vi V2)t 1 2 1 2 123 15154 4 3 4 3 1t t t t tt t2 = 15 . 4 = 60 pht = 1 giVy qung ng CB l 3km, AB = 15kmNgi xut pht t 11 gi 45 pht (15:4) = 8 giCu 4e. Tam gic AIB = tam gic CID v c (IB = ID;gc I1 = gc I2; IA = IC)f. Tam gic AID = tam gic CIB (c.g.c) gc B1 = gc D1 v BC = AD hay MB =ND tam gic BMI = tam gic DNI (c.g.c) Gc I3 = gc I4 M, I, N thng hng v IM = INDo vy:I l trung im ca MNg. Tam gic AIB c gc BAI > 900 gc AIB < 900 gc BIC > 900h. Nu AC vung gc vi DC th AB vung gc vi AC do vy tam gic ABC vung ti ACu 5.P = 4 10 1014 4xx x + + P ln nht khi 104 x ln nhtTrang 123Nm hc 2010-2011Xt x > 4 th104 x < 0Xt x< 4 th104 x > 0104 x ln nht 4 x l s nguyn dng nh nht 4 x = 1 x = 3khi 104 x = 10 Pln nht = 11.-------------------------------------------------------------H ng dn chm 26 Bi 1 :a) Tm x . Ta c 6 2 x + 5x =96 2 x = 9-5x* 2x 6 0 x 3 khi 2x 6 = 9-5x x = 715khng tho mn. (0,5)* 2x 6 < 0x< 3 khi 6 2x = 9-5x x= 1tho mn. (0,5)Vy x = 1.b) Tnh . (1+2+3+...+90).( 12.34 6.68) : ,_+ + +61514131 = 0. (0,5)( v 12.34 6.68 = 0).c) Ta c : 2A = 21 + 22 +23 + 24 + 25 +...+ 2101 2A A = 2101 1. (0,5)Nh vy 2101 1 < 2101 . Vy A1 . A = 5 tc l 4923511 +x xxx. (1)Bi 4 : E thuc phn gic ca ABCnn EN = EC ( tnh cht phn gic) suy ra :tam gic NEC cn v ENC = ECN (1) . D thuc phn gic ca gc CAB nn DC = DM(tnh cht phn gic ) suy ra tam gic MDC cn .v DMC =DCM,(2) . Ta li c MDB = DCM +DMC (gc ngoi ca CDM ) = 2DCM.Tng t ta li c AEN = 2ECN . M AEN = ABC (gc c cnh tng ng vung gc cng nhn).Trang 125Nm hc 2010-2011MDB = CAB (gc c cnh tng ng vung gc cng nhn ). Tam gic vung ABC cACB = 900 , CAB + CBA = 900 , suy ra CAB = ABC = AEN + MDB = 2 ( ECN + MCD )suy raECN + MCD = 450 . Vy MCN = 900 450 =450 . (1,5)Bi 5 :Ta c P = -x2 8x + 5 = - x2 8x 16 +21 = -( x2 +8x + 16) + 21 = -( x+ 4)2 + 21; (0,75)D