- 1. www.VNMATH.com S GIO DC BNH NHK THI TUN SINH VO LP 10 BNH
NHTRNG THPT CHUYN L QU N NM HC 2009-2010 chnh thc Mn thi:Ton
(chuyn) Ngy thi:19/06/2009 Thi gian:150 pht Bi 1(1.5im) Cho a,b,c l
di ba cnh ca mt tam gic.Chng minh rng: ab c 1 2 bc c a a b Bi
2(2im) 111 Cho 3 s phn bit m,n,p.Chng minh rng phng trnh 0 c haixm
xn x p nghim phn bit. Bi 3(2im) 111 Vi s t nhin n, n 3 .t Sn ... 3
1 2 5 2 3 2n 1 n n 1 1 Chng minhSn< 2 Bi 4(3im) Cho tam gic ABC
ni tip trn tm O c di cc cnh BC = a, AC = b, AB = c.E l im nm trn
cung BC khng cha im A sao cho cung EB bng cung EC.AE ct cnh BC ti
D. a.Chng minh:AD2 = AB.AC DB.DC b.Tnh di AD theo a,b,c Bi
5(1.5im)m 1 Chng minh rng : 2 Vi mi s nguyn m,n.n n 2 3 2
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2. www.VNMATH.comP N MN TON THI VO 10 TRNG CHUYN L QU N NM 2009
Bi 1: V a,b,c l di ba cnh tam gic nn ta c:a,b,c >0 v a< b+c
,b< a + c , c < a+b a aa2a Nn ta c bc a bc ab caa Mt khc bc a
bc aa2a Vy ta c (1) abc cb abc b b2b cc 2a Tng t (2); (3) abc ca
abcabc ba abc Cng (1) (2) v (3) v theo v ta c iu phi chng minh. Bi
2: K: x m, n, p PT cho (x-n)(x-p)+(x-m)(x-p)+(x-m)(x-n) = 0 3x2
-2(m+n+p)x +mn+mp+np = 0(1) Ta c ( m n p )2 3(mn mp np ) = m2+n2+p
2 +2mn+2mp+2np -3mn-3mp-3np = 1 m2+n2+p2 mn-mp-np =
[(m-n)2+(n-p)2+(m-p)2] >0 2 2 t f(x) = 3x -2(m+n+p)x + mn+ mp
+np Ta c f(m) = 3m2 2m2 -2mn -2mp +mn +mp +np = m2 mn mp +np =
(m-n)(m-p) 0 = >m,n,p khng phi l nghim ca pt(1) Vy PT cho lun c
hai nghim phn bit Bi 3 1n 1 nn 1 n Ta c : 2n 1 n n 1 2n 1 4n2 4n 1n
1 n n +1 - n 1 11 4n 4n 22 n 1. n 2 n n 11 1 1 1 1 1 1 1 1 1 Do Sn
1 ... 2 22 2 3 nn 1 n 1 2 Bi 3:C Ta c BAD CAE ( Do cung EB = cung
EC) V AEC DBA ( Hai gc ni tip cng chn cung AC) nn BAD EAC BA AEEa
AB.AC AE. AD(1)Ob AD AC Ta c ADC BDC (i nh) v CAD DBE D (2 gc ni
tip cng chn cung CE) nn ACD BDE AD DB AD.DE DB.DChay DC DE cA
AD(AE-AD) = DB.DC B Hay AD2 = AD.AE - DB.DC=AB.AC DB.DC (do (1)) DC
DB DC DB DC DBa 4b)Theo tnh cht ng phn gic ta c hay AC ABbcb c
bcwww.VNMATH.com2 3. www.VNMATH.com DC DBa aa 2 bc vy . . DB.DC b
cbc bc b c2 a 2 bca2 theo cu a ta c AD2 = AB.AC DB.DC = bc bc 1 b c
2 b c2 a2 AD bc 1 b c2 Bi 5:mm V l s hu t v 2l s v t nn 2 n n Ta
xet hai trng hp:m a) 2 Khi m 2 2 n 2 m2 2 n 2 1 hay m 2n 2 1n T suy
ra : 1 2 2 2m2n2 1 1 n1 1 2 2 2 2 2 nnn1 21 n2 2 2 2 n 2 2 2 3 2 n
nm b) 2 Khi m 2 2n 2 m 2 2n 2 1 hay m 2n 2 1n T suy ra : 1 22 m m
2n2 11 n2 2 2 2 2 2 2 n n n n 1 2 2n211 n2 2 2 2 1 n 2 3 2 n
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www.VNMATH.comS GD&T VNH PHC K THI VO LP 10 THPT CHUYN NM HC
2009-2010 THI MN: TON Dnh cho cc th sinh thi vo lp chuyn Ton CHNH
THC Thi gian lm bi: 150 pht, khng k thi gian giao ( c 01 trang) Cu
1: (3,0 im)1 1 9x y x y 2 a) Gii h phng trnh: xy 1 5xy 2 b) Gii v
bin lun phng trnh: | x 3 | p | x 2 | 5 (p l tham s c gi tr thc). Cu
2: (1,5 im)Cho ba s thc a, b, c i mt phn bit. a2 b2 c2Chng minh 2(b
c)2 (c a) 2 ( a b)2 Cu 3: (1,5 im)1 2x 2 Cho A v B 24x 4 x 1x2 2x
12A B Tm tt c cc gi tr nguyn ca x sao cho C l mt s nguyn.3 Cu 4:
(3,0 im) Cho hnh thang ABCD (AB // CD, AB x3 x1 1; x4 x2 1 x1 x2
b(1) x . x c(2) 1 2 Theo h thc Vi t ta c 2 x1 1 x2 1 b (3) x 1 . x
1 bc(4) 122 T (1 ) v ( 3 ) => b + b - 2 = 0 b = 1 ; b = -2 t ( 4
) => x1 . x2 x1 x2 1 bc => c - b + 1 = bc ( 5 ) +) vi b = 1
th ( 5 ) lun ng , phng trnh x2 + +b x + c = 0 tr thnh 1 X2 + x + 1
= 0 c nghim nu 1 4 c 0 c 4 +) vi b = -2 ( 5 ) tr thnh c + 3 = -2 c
=> c = -1 ; phng trnh x2 + b x + c = 0 tr thnh x2 - 2 x - 1 = 0
c nghim l x = 1 2 1 vy b= 1; c c ; 4 b = -2 ; c = -1 Bi 3 : 1. p
dng bt ng thc C si cho 3 s dng 1 1 11 a b c 3 abc 33 a b c abc 1 1
1 => a b c 9a b c du = sy ra a = b = c 2 222 2. ta c ab bc ca a
b c ab bc ca a b c 3 3 2007 669 ab bc cawww.VNMATH.com8 9.
www.VNMATH.com p dng cu 1 ta c 11 1 2 a b c 2ab 2bc 2ca 92 2 222 a
b cab bc ca ab bc ca 119 => 2 1 2 a b c 2 ab bc ca a b c 21 2009
vy 2 22 670 . du = sy ra a = b = c = 1a b c ab bc ca 1 A BOP BAO
ABO B 2 0 1 180 C BA Bi 4 : a) ta c PNC 22 BOP PNC => t gic BOPN
ni tip +) tng t t gic AOQM ni tip +) do t gic AOQM ni tip=> AMO
900 AQO t gic BOPN ni tip => BPO BNO 900 => AQB APB 900 =>
t gic AQPB ni tip b ) tam gic AQB vung ti Qc QE l trung tuyn nn QE
= EB = EA 1 => EQB EBQ B QBC => QE //BC 2 M E F l ng trung
bnh ca tam gic ABC nn E F //BC Q; E; F thng hng c)MP OM OP MOP ~
COB( g g ) a OC OBNQ ON OM NOQ ~ COA( g g ) bOC OCPQ OP OM POQ ~
BOA( g g ) cOB OC OM MP NQ PQ MP NQ PQ OCabcA B C Bi 5 : 1) 3x - y3
= 1 y 1 3m y 3m 1 3x y 1 y 2 y 1 => tn ti m; n sao cho y 2 y 1
3n 9 m 3.3m 3 3nm b x m b x +) nu m = 0 th y = 0 v x = 0 9 m 3.3m 3
3 3n 3 +) nu m > 0 th m n n 1 9 3.3 3 9 3 9m => 9 m 3.3m 3 3
3m 3m 3 0 => m = 1 => y = 2 ; x = 2 vy p/ trnh c hai nghim l
( 0 ; 0 0 ; ( 2 ; 2 ) 2.Ta t mu cc vung ca bng bng hai mu en trng
nh bn c vua Lc u tng s si cc en bng 1005 . 2009 l mt s l sau mi php
thc hin thao tc T tng s si cc en lun l s lwww.VNMATH.com9 10.
www.VNMATH.com vy khng th chuyn tt c vin si trn bng vung v cng mt
sau mt s hu hn cc php thc hin thao tc T S gio dc-o to K thi tuyn
sinh vo lp 10 THPT chuyn H namNm hc 2009-2010 Mn thi : ton( chuyn)
chnh thc Thi gian lm bi: 120 pht(khng k thi gian giao ) Bi 1.(2,5
im)1 1 1) Gii phng trnh:2 2 x 3x 2 x 21x x y 7 2) Gii h phng trnh:
x 12x y Bi 2.(2,0 im)Cho phng trnh: x 6 x 3 2m 0 a) Tm m x = 7 48 l
nghim ca phng trnh. b) Tm m phng trnh c 2 nghim x=x1; x=x2 tho mn:
x1 x2 24 x1 x23 Bi 3.(2,0 im)1) Cho phng trnh: 2 x 2 2 2m 6 x 6 m
52 0 ( vi m l tham s, x l n s). Tmgi tr ca m l s nguyn phwowng trnh
c nghim l s hu t.22) Tm s abc tho mn: abc a b 4c .Bi 4.(3,5 im) Cho
ABC nhn c C A. ng trn tm I ni tip ABC tip xc vi cc cnhAB, BC, CA ln
lt ti cc im M, N, E; gi K l giao im ca BI v NE. C a) Chng minh: AIB
900 .2b) Chng minh 5 im A, M, I, K, E cng nm trn mt ng trn.c) Gi T
l giao im ca BI vi AC, chng minh: KT.BN=KB.ET.www.VNMATH.com10 11.
www.VNMATH.comd) Gi Bt l tia ca ng thng BC v cha im C. Khi 2 im A,
B v tia Bt c nh; im C chuyn ng trn tia Bt v tho mn gi thit, chng
minh rng cc ng thng NE tng ng lun i qua mt im c nh. -----------
Ht---------- H v tn th sinh:..S bo danh: Ch k gim th s 1:.Ch k gim
th s 2..Gi mt s cu kh trong thi: Bi 3:21) Ta c = 4m 2 12 m 68 2m 3
77 phng trnh c nghim hu t th phi l s chnh phng. Gi s= n2( trong n l
s t nhin).Khi ta c 22 2m 3 77 n2 2m 3 n2 77 2m 3 n . 2m 3 n 77Do n
N nn 2m-3+n>2m-3-nV do m Z, n N v
77=1.77=7.11=-1.(-77)=-7.(-11)T xt 4 trng hp ta s tm c gi tr ca
m.2)T gi thit bi ton ta c: 2100 a 10b 2 100a 10b c a b .4c c 2 (do
4 a b 1 0)4 a b 110 10 a b 10 a b 9a 224 a b 14 a b 122 Ta c 4 a b
1 l s l v do 0 c 9 nn 4 a b 1 5.2 22M 4 a b l s chn nn 4 a b phi c
tn cng l 6 a b phi c tn cng l 4 hoc 9. (*) 2.5 abMt khc c v 4( a b)
2 12 2 2 4 a b 1 l s l 4 a b 1 6 V A, B la giao iem cua (d) va (P)
nen hoanh o xA, xB phai thoa man pt: mx2 + 9x 9m = 0 9 Theo Vi-et
ta co: xA+ xB =; xA. xB = -9 m Do A, B 3 3 (d ) y A x A 3 ; yB xB 3
mm Theo cong thc tnh khoang cach: 2 2AB x A xB y A y B 2 2 3 3 x A
xB xA xB m m 2 9 2 x A xB 2 Ax xB m 2 9 x A xB 1 2 m 2 9 x A x B 4
x A . xB 1 2 m 9 2 9 4(9) 1 2 m m 81 9 2 36 1 2 m m 81 729 324 36
36 6m2 m 4 m 2 Bai 5: (3 iem) Cho hai ng tron (O ; R) va (O ; R)
cat nhau tai A va B (R > R). Tiep tuyen tai B cua (O ; R) cat (O
; R) tai C va tiep tuyen tai B cua (O ; R) cat (O ; R) tai D. a/
Chng minh rang: AB2 = AC.AD va2 BC AC BD AD b/ Lay iem E oi xng cua
B qua A. Chngwww.VNMATH.com15 16. www.VNMATH.com minh bon iem B, C,
E, D thuoc mot ng tron co tam la K. Xac nh tam K cua ng tron. a/
Xet (O) ta co C1 B2 (chan cung AnB) Xet (O) ta co D B (chan cung
AmB) 11 ABC ADBAB AC BC(1)AD AB BDE AB 2 AC . AD//2 1C2 2 BC AB AB
2 AC. AD AC 21= 2D BD AD AD 2 AD 2 AD x1 21K A b/ T (1) thay AE =
AB ta co O =OAE AC x (*) mat khac: 12AD AE j C B ; A B D A1 B112 2
1 A A (**)1 2 T (*) va (**) suy ra: AEC ADE (c g c) E D2 2 CED CBD
E1 E2 B1 B2 E D D B1 2 120 180 ( xet BDE ) Vay t giac BCED noi tiep
ng tron tam K. Vi K la gaio iem 3 ng trc cua BCE hoac
BDEwww.VNMATH.com 16 17. www.VNMATH.comS GD&T Ngh An K thi TUYN
sinh VO lp 10trng thpt chuyn phan bi chu thi chnh thc nm hc 2009 -
2010 Mn thi: TON Thi gian: 150 pht, khng k thi gian giao Bi 1: (3.5
im) a) Gii phng trnh3x2 3 7x 3b) Gii h phng trnh 8 2 3x y 3 x3 2 6
y Bi 2: (1.0 im) Tm s thc a phng trnh sau c nghim nguyn x 2 ax a 2
0 . Bi 3: (2.0 im) Cho tam gic ABC vung ti A c ng phn gic trong BE
(E thuc AC). ng trn ng knh AB ct BE, BC ln lt ti M, N (khc B). ng
thng AM ct BC ti K. Chng minh: AE.AN = AM.AK. Bi 4: (1.5 im) Cho
tam gic ABC c 3 gc nhn, trung tuyn AO c di bng di cnh BC. ng trn ng
knh BC ct cc cnh AB, AC th t ti M, N (M khc B, N khc C). ng trn
ngoi tip tam gic AMN v ng trn ngoi tip tam gic ABC ct ng thng AO ln
lt ti I v K. Chng minh t gic BOIM ni tip c mt ng trn v t gic BICK l
hnh bnh hnh. Bi 5: (2.0 im)a) Bn trong ng trn tm O bn knh 1 cho tam
gic ABC c din tch ln hn hoc bng 1. Chng minh rng im O nm trong hoc
nm trn cnh ca tam gic ABC.b) Cho a, b, c l cc s thc dng thay i tha
mn: a b c 3 . Tm gi tr nh nht ca biu thc ab bc ca P a 2 b2 c 2 a 2b
b 2c c 2
a----------------------------------------Ht----------------------------------------H
v tn th sinh .... SBD..www.VNMATH.com17 18. www.VNMATH.com * Th
sinh khng c s dng ti liu. * Gim th khng gii thch g
thm.www.VNMATH.com18 19. www.VNMATH.com S GD&T Ngh An K thi
TUYN sinh VO lp 10 trng thpt chuyn phan bi chu nm hc 2009 - 2010
thi chnh thc Mn thi: Ton Hng dn chm thiBn hng dn chm gm 03 trang Ni
dung p n imBi 13,5 a 2,03x2 3 7x 3 x 2 7 x 3 3 x 2. 3 7 x 3x 2 3 7
x 27 0.50 9 9. 3 ( x 2)(7 x) 27 0.25 3 ( x 2)(7 x ) 20.25 ( x 2)(7
x) 8 0.252 x 5x 6 00.25 x 1 ( tha mn ) 0.50 x 6 b 1,502 t z 0.25y2
3 x z 3 H cho tr thnh 0.25 32 3 z x 3 x z z 3 x3 0,25 x z x 2 xz z
2 3 00,25 22 xz (v x xz z 3 0, x, z ).0,25 x 1 T ta c phng trnh: x
3 3 x 2 0 x 20,25 Vy h cho c 2 nghim: ( x, y ) ( 1; 2), 2,1Bi 2:
1,0 2 iu kin phng trnh c nghim: 0 a 4a 8 0 (*).0,25 Gi x1, x2 l 2
nghim nguyn ca phng trnh cho ( gi s x1 x2). x1 x2 a 0,25 Theo nh l
Viet: x1.x2 x1 x2 2 x1.x2 a 2 ( x1 1)( x2 1) 3 x 1 3 x1 1 1 1 hoc
(do x1 - 1 x2 -1) x2 1 1 x2 1 3 x1 4 x1 0 hoc x2 2 x2 2
0,25www.VNMATH.com 19 20. www.VNMATH.com Suy ra a = 6 hoc a = -2
(tha mn (*) ) Th li ta thy a = 6, a = -2 tha mn yu cu bi ton.
0,25Bi 3:2,0 V BE l phn gic gc nn MBC MNABC ABM AM 0,25 MAE MAN (1)
0,50A V M, N thuc ng trn ng0,25 knh AB nn 900 AMB ANB 90 , kt hp0E
ANK AME vi (1) ta c tam gic AME ng0,50M dng vi tam gic ANK AN
AK0,25 BCAM AEN K 0,25 AN.AE = AM.AK (pcm)Bi 4:1,5 V t gic AMIN ni
tip nn ANM AIMV t gic BMNC ni tip nn A 0,25 ANM ABC .Suy ra t gic
BOIM ni tipAIM ABCT chng minh trn suy ra tam gic AMIE ng dng vi tam
gic AOBN 0,25AM AIM AI . AO AM . AB (1)AO ABI Gi E, F l giao im ca
ng thng AOB C vi (O) (E nm gia A, O).OChng minh tng t (1) ta c:K
0,25AM.AB = AE.AF = (AO - R)(AO + R) (vi BC = 2R) = AO2 - R2 =
3R2F3R 2 3 R 2 3R R AI.AO = 3R2 AI OI (2)0,25AO 2 R2 2 Tam gic AOB
v tam gic COK ng dng nn OA.OK = OB.OC = R2R2 R 2 R0,25 OK (3)OA 2 R
2 T (2), (3) suy ra OI = OK Suy ra O l trung im IK, m O l trung im
ca BC 0,25 V vy BICK l hnh bnh hnhBi 5:2,0 a,1,0 A Gi s O nm ngoi
min tam gic ABC.Khng mt tnh tng qut, gi s A v O 0,25nm v 2 pha ca
ng thng BCK Suy ra on AO ct ng thng BC ti K.B C 0,25 HK AH vung gc
vi BC ti H.OSuy ra AH AK < AO 0 2 ab bc ca22 Suy ra P a b c a 2
b2 c20,25 222 9 (a 2 b 2 c 2 )P a b c 2(a 2 b 2 c 2 ) t t = a2 + b2
+ c2, ta chng minh c t 3. 9t t 9 t 13 1 Suy ra P t 3 4 P 42t2 2t 2
22 2 0,25 Du bng xy ra khi v ch khi a = b = c = 1 Vy gi tr nh nht
ca P l 4Nu th sinh gii cch khc ng ca mi cu th vn cho ti a im ca cu
S GIO DC V O TO K THI TUYN SINH VO LP 10 THPT CHUYN LAMSN THANH HO
NM HC: 2009-2010 chnh thcMN: TON (Dnh cho hc sinh thi vo lp chuyn
Ton) Thi gian: 150 pht (khng k thi gian giao )Ngy thi: 19 thng 6 nm
2009 Cu 1: (2,0 im) 11. Cho s x ( x R ; x > 0 ) tho mn iu kin :
x 2 += 7 . Tnh gi tr cc biu x2 11thc : A = x 3 + 3 v B = x 5 + 5 .
x xwww.VNMATH.com21 22. www.VNMATH.com 11 + 2- 2 xy 2. Gii h phng
trnh: 1 + 2- 1 2 yxCu 2: (2,0 im)Cho phng trnh: ax2 + bx + c = 0 (a
0) c hai nghim x1, x2 tho mn iu kin:2a 2 - 3ab + b20 x1 x 2 2 . Tm
gi tr ln nht ca biu thc: Q =. 2a 2 - ab + acCu 3: (2,0 im) 1 1. Gii
phng trnh:x-2 +y + 2009 + z - 2010 = x + y + z . 2 2. Tm tt c cc s
nguyn t p 4p2 + 1 v 6p2 + 1 cng l s nguyn t.Cu 4: (3,0 im) 1. Cho
hnh vung ABCD c hai ng cho ct nhau ti E. Mt ng thng i qua A, ct cnh
BC ti M v ct ng thng CD ti N. Gi K l giao im ca cc ng thng EM v BN.
Chng minh rng: CK BN. 2. Cho ng trn (O) bn knh R = 1 v mt im A sao
cho OA = 2 . V cc tip tuyn AB, AC vi ng trn (O) (B, C l cc tip im).
Mt gc xOy c s o bng 45 0 c cnh Ox ct on thng AB ti D v cnh Oy ct on
thng AC ti E. Chng minh rng2 2 - 2 DE < 1.Cu 5: (1,0 im)Cho biu
thc P = a2 + b 2 + c2 + d2 + ac + bd , trong ad bc = 1. Chng minh
rng: P 3.-------------------------------------------------- Ht
---------------------------------------------------H v tn th sinh:
.. S bo danh: ..s gio dc - o to hk thi tuyn sinh vo lp 10 thpt
chuynnam Nm hc 2009 - 2010 Mn thi : ton( chung) chnh thcThi gian lm
bi: 120 pht (Khng k thi gian giao )Bi 1. (2 im) 2 Cho biu thc P =x
x 1 x 2 3 x x 1 x1 x a) Tm iu kin xc nh ca P b) Rt gn P c) Tm x P
> 0Bi 2. (1,5 im)www.VNMATH.com 22 23. www.VNMATH.comGii h phng
trnh: 1 2 x y 2 2 2 x y 1 Bi 3. (2 im) 1) Tm to giao im ca ng thng
y = x + 6 v parabol y = x2 2) Tm m th hm s y = (m + 1)x + 2m + 3 ct
trc , trc Oy ln lt ti cc im A , Bv AOB cn ( n v trn hai trc v Oy
bng nhau).Bi 4. (3,5 im)Cho ABC vung nh A, ng cao AH, I l trung im
ca Ah, K l trung im ca HC.ng trn ng knh AH k hiu (AH) ct cc cnh AB,
AC ln lt ti dim M v N. a) Chng minh ACB v AMN ng dng b) Chng minh
KN l tip tun vi ng trn (AH) c) Tm trc tm ca ABKBi 5. (1 im)Cho x,
y, z l cc s thc tho mn: x + y + x = 1. 1 1 1Tm gi tr nh nht ca biu
thc P = 16 x 4 y z---------ht---------H v tn th sinh:..S bo
danh:....Ch k gim th s 1: Ch k gim th s 2:..s gio dc o to h K thi
tuyn sinh vo lp 10 thpt chuyn nam Nm hc 2009 2010hng dn chm thi mn
ton : chungBi 1 (2 im)a) (0,5 im) iu kin xc nh ca P l x 0 v x 1
0.5b) (1 im)x x 1 x0,251 x1 x2 x 2 3 x xx4 x 43 x x0,251 x 1 x 4
x0,25 1 x4Vy P =0,251 xc) (0,5 im) P>0 1 x 00,25 x 1 0 x 1
0,25Bi 2 (1,5 im)Cng hai phng trnh ta c : 3 2 2 x 1 2 0,51 2 1x 2 1
0,5 3 2 2 1 2 www.VNMATH.com23 24. www.VNMATH.comVi x 2 1 y 2 2 1 2
1 1 2 10,25x 2 1K/l Vy h c nghim: 0,25 y 2 1Bi 3 (2 im)a) (1 im)
Honh giao im l nghim ca phng trnh: x2 = x + 6 05 x 2 x 6 0 x 2 hoc
x = 3Vi x = -2 y 4; x 3 y 90,25Hai im cn tm l (-2;4); (3;9)0,25b)
(1 im) 2m 3 2m+3 Vi y = 0 m 1 2m 3 0 x (vi m -1) A - ;0 0,25m 1 m+1
Vi x = 0 y 2m 3 B 0;2m+32m 3 OAB vung nn OAB cn khi A;B O v OA = OB
2m 30,25 m 1 2m 3 1 3+ Vi 2m 3 2m 3 1 0 m 0 hoc m = (loi) 0,25m 1 m
1 2 2m 3 1 3+ Vi 2m 3 2 m 3 1 0 m 2 hoc m = (loi)m 1 m 1 20,25K/l:
Gi tr cn tm m = 0; m = -2Bi 4(3,5 im)a) (1,5 im)A N 0,25 EIMCBH K
AMN v ACB vung nh A0,25 C AMN AHN (cng chn cung AN) AHN ACH (cng ph
vi HAN ) (AH l ng knh)0,75 AMN ACH AMN ACB0,25 b) (1 im) HNC vung
nh N v ANH 900 c KH = KC NK = HKli c IH = IN (bn knh ng trn (AH)) v
IK chung nn KNI = KHI (c.c.c)0,75 KNI KHI 900 KNI 900C KN In, IN l
b knh ca (AH) KN l tip tuyn vi ng trn (AH)0,25c) (1 im)+ Gi E l
giao im ca Ak vi ng trn (AH), chng minh gc HAK= gc HBIHA HK0,5 Ta c
AH2 HB.HC AH.2IH = HB.2HK HB HI www.VNMATH.com 24 25.
www.VNMATH.com HAK HBI HAK HBI + C HAK EHK (chn cung HE) HBI EHK BI
// HE0,25C 900 (AH l ng knh) BI AK AEH ABK c BI AK v BK AI I l trc
tm ABK 0,25Bi 5 (1 im) 1 1 1 11 1 yx z x z y 21 0,5P= x y z 16x 4 y
z 16x 4 y z 16 x 4 y 16 x z 4 y z 16y x 1Theo ci vi cc s dng: du
bng xu ra khi y=2x16 x 4 y 4zx 1 du bng xu ra khi z=4x16 x z 20,25
z y 1 du bng xu ra khi z=2y4y zVy P 49/16P = 49/16 vi x = 1/7; y =
2/7; z = 4/70,25Vy gi tr b nhy ca P l 49/16S GIO DC V O TO THI TUYN
SINH VO LP 10 CHUYNTNH NINH BNH NM HC 2009 2010 Mn Ton Vng 1 CHNH
THC(Dng cho tt c cc th sinh)Thi gian lm bi 120 pht (Khng k thi gian
giao ) thi gm 05 cu trong 01 trang Cu 1: (2 im) Tnh gi tr biu thc:
x 5 22 5 5 25033y 3 1 3 1 x xy yA x xy y x y Cu 2: (2,5 im) Cho
phng trnh (m + 1)x2 2(m 1) + m 2 = 0 (n x, tham s m). a) Gii phng
trnh khi m = 2.1 1 7b) Tm m phng trnh c hai nghim phn bit x1; x 2
tha mn: x1 x 2 4 Cu 3: (1,0 im)Khong cch gia hai bn sng A v B l 60
km. Mt ca n chy xui dng t bn A ti bn B, ngh 1 gi 20 pht bn sng B v
ngc dng tr v A. Thi gian k t lc khi hnh n khi v bn A tt c 12 gi.
Tnh vn tc ring ca ca n v vn tc dng nc bit vn tc ring cu ca n gp 4
ln vn tc dng nc. Cu 4: (3,5 im) www.VNMATH.com25 26.
www.VNMATH.comCho ng trn (O; R) v ng thng (d) khng i qua tm O ct ng
trn (O;R) ti hai im phn bit A, B. im M chuyn ng trn (d) v nm ngoi
ng trn (O;R), qua M k hai tip tuyn MN v MP ti ng trn (O; R) (N, P l
hai tip im).a) Chng minh rng t gic MNOP ni tip c trong mt ng trn,
xc nhtm ng trn .b) Chng minh MA.MB = MN2.c) Xc nh v tr im M sao cho
tam gic MNP u.d) Xc nh qu tch tm ng trn ngoi tip tam gic MNP.Cu 5:
(1 im)4 5Cho hai s thc dng x, y tha mn: 23x y67Tm gi tr nh nht ca
biu thc: B 8x 18y xywww.VNMATH.com 26 27. www.VNMATH.comp n:Cu 1:x
= 10;y=3A=xy=7Bi 2:a) Vi m = 2 th x1 = 0; x 2 = 2/3.b) m = -6.Bi
3:S: Vn tc ca n: 12 km/hVn tc dng nc: 3 km/hBi 4:a, b).c) Tam gic
MNP u khi OM = 2Rd) Qu tch tm ng trn ngoi tip tam gic MNP l ng thng
d song song ving thng d (tr cc im bn trong ng trn).Bi 5: 6 7B 8x
18y x y2 2 4 5 8x 18y 8 12 23 43x y x y 1 1Du bng xy ra khi x; y ;
. 2 31 1Vy gi tr nh nht ca B l 43 khi x; y ; 2 3www.VNMATH.com 27
28. www.VNMATH.comS GIO DC V O TO THI TUYN SINH TRUNG HC PH THNG
TNH PH YN NM HC 2009-2010Mn thi: TON CHUYN CHNH THCThi gian: 150
pht (khng k thi gian pht ) *****Cu 1.(4,0 im) Cho phng trnh x4 +
ax3 + x2 + ax + 1 = 0, a l tham s . a) Gii phng trnh vi a = 1. b)
Trong trng hp phng trnh c nghim, chng minh rng a2 > 2.Cu 2.(4,0
im) a) Gii phng trnh: x + 3 + 6 - x (x + 3)(6 - x) = 3 .x + y + z
=1 b) Gii h phng trnh: . 2x + 2y - 2xy + z 2 = 1Cu 3.(3,0 im) Tm tt
c cc s nguyn x, y, z tha mn : 3x2 + 6y2 +2z2 + 3y2z2 -18x = 6.Cu
4.(3,0 im) a) Cho x, y, z, a, b, c l cc s dng. Chng minh rng: 3 abc
+ 3 xyz 3 (a + x)(b + y)(c + z) . 3 b) T suy ra : 3 3 3 3 3 3 3 23
3Cu 5.(3,0 im) Cho hnh vung ABCD v t gic MNPQ c bn nh thuc bn cnh
AB, BC,CD, DA ca hnh vung. ACa) Chng minh rng SABCD (MN + NP + PQ +
QM).4b) Xc nh v tr ca M, N, P, Q chu vi t gic MNPQ nh nht.Cu 6.(3,0
im) Cho ng trn (O) ni tip hnh vung PQRS. OA v OB l hai bn knh thayi
vung gc vi nhau. Qua A k ng thng Ax song song vi ng thng PQ, qua B
kng thng By song song vi ng thng SP. Tm qu tch giao im M ca Ax v
By. =HT=H v tn th sinh:.S bo danh:Ch k gim th 1:Ch k gim th
2:.www.VNMATH.com28 29. www.VNMATH.comS GD & T PH YN***K THI
TUYN SINH THPT NM HC 2009 -2010MN : TON (H s 2)------- CHNH THCHNG
DN CHM THI Bn hng dn chm gm 04 trangI- Hng dn chung:1- Nu th sinh
lm bi khng theo cch nu trong p n m vn ng th cho imtng phn nh hng dn
quy nh.2- Vic chi tit ho thang im (nu c) so vi thang im hng dn chm
phi bo mkhng sai lch vi hng dn chm v c thng nht thc hin trong Hi ng
chm thi.3- im ton bi thi khng lm trn s.II- p n v thang im:CU P N im
432Cu 1a. Ta c phng trnh : x + ax +x + ax + 1 = 0 (1) (2,0) Khi a
=1 , (1) x 4 +x 3 +x 2 +x+1= 0(2) D thy x = 0 khng phi l nghim. 11
Chia 2 v ca (2) cho x2 ta c: x 2 + 2 + x + +1= 0 (3). 0,50 xx 1 1
11 t t = x+ t x+ x + 2 v x 2 + 2 t 2 -2 . x x x x0,50 Phng trnh (3)
vit li l : t 2 + t - 1 = 0 1 5 1 5 0,50 Gii (3) ta c hai nghim t1 v
t 2 u khng tha iu 2 2 kin |t| 2.Vy vi a = 1, phng trnh cho v nghim.
0,50Cu1b.V x = 0 khng phi l nghim ca (1) nn ta cng chia 2 v cho x2
ta c(2,0)11 phng trnh : x 2 + 2 +a x + +1= 0 . 0,50 x x1 t t = x +
, phng trnh s l : t2 + at - 1 = 0 (4).xDo phng trnh cho c nghim nn
(4) c nghim |t| 2. T (4) suy ra0,501- t 2a . t 0,50 (1 - t 2 )2T :
a 2 >2 2 t 2 (t 2 - 4) 1 0 (5)0,50t2V |t| 2 nn t2 >0 v t2 4 0
, do vy (5) ng, suy ra a2 > 2.Cu 2a. x + 3 + 6 - x - (x + 3)(6 -
x) 3 (1) (2,0) x+3 0iu kin : -3 x 6 . 6-x 0 u x + 30,50 t :, u , v
0 u 2 v 2 9. v = 6 - x 0,50Phng trnh c tr thnh h :www.VNMATH.com 29
30. www.VNMATH.com u 2 + v2=9 (u + v) 2 - 2uv = 9 0,50 u + v - uv =
3u + v= 3 + uv uv = 0 u = 0 Suy ra :(3+uv)2-2uv = 9 uv = -4v = 0
0,50 x+3 = 0 x = -3 . 6-x = 0 x = 6 Vy phng trnh c nghim l x =-3 ,
x = 6. Cu Ta c h phng trnh : 2b. x+y+z=1 x+y = 1-z(2,0)2 2 0,50
2x+2y-2xy+z =1 2xy = z +2(x+y)-1 x + y = 1 - z22 0,50 2xy = z - 2z
+ 1 = (1- z) 2xy = (x + y)2 x 2 + y2 = 0 x = y = 0 z = 1 .0,50 Vy h
phng trnh ch c 1 cp nghim duy nht: (x ;y ;z) = (0 ;0; 1). 0,50Cu
3.(3,0) Ta c : 3x2 + 6y2 + 2z2 +3y2z2 -18x = 6 (1) 0,50 3(x-3)2 +
6y2 + 2z 2 + 3y2 z 2 33 (2) Suy ra : z2 3 v 2z2 33 0,50 Hay |z| 3.
V z nguyn suy ra z = 0 hoc |z| = 3. a) z = 0 , (2) (x-3)2 + 2y2 =
11 (3) 0,50 T (3) suy ra 2y2 11 |y| 2.Vi y = 0 , (3) khng c s nguyn
x no tha mn. 0,50Vi |y| = 1, t (3) suy ra x { 0 ; 6}. b) |z| = 3,
(2) (x-3)2 + 11 y2 = 5 (4)0,50 T (4) 11y2 5 y = 0, (4) khng c s
nguyn x no tha mn. Vy phng trnh (1) c 4 nghim nguyn (x ;y ;z) l
(0;1;0) ; (0 ;-1;0) ;0,50 (6 ;1 ;0) v (6 ;-1 ;0).Cu 4a. 3 abc 3 xyz
3 (a+x)(b+y)(c+z) (1) (2,0) Lp phng 2 v ca (1) ta c : 0,50 abc +
xyz + 3 3 (abc)2 xyz +3 3 abc(xyz)2 (a+x)(b+y)(c+z) abc + xyz+ 3 3
(abc)2 xyz + 3 3 abc(xyz)2 abc+xyz+abz+ayc+ayz+xbc+xyc+xbz 0,503 2
32 3 (abc) xyz +3 abc(xyz) (abz+ayc+ xbc)+ (ayz+xbz+xyc) (2) Theo
bt ng thc Cauchy, ta c : (abz+ayc+ xbc) 3 3 (abc) 2 xyz (3)0,5032
(ayz+xbz+ xyc) 3 abc(xyz) (4) 0,50 Cng hai bt ng thc (3) v (4) ta c
bt ng thc (2), do (1) c chng minh.www.VNMATH.com 30 31.
www.VNMATH.comCu4b. p dng BT (1) vi a = 3+ 3 3, b = 1, c = 1, x = 3
- 3 3, y = 1, z = 1(1,0) 3 0,50 Ta c : abc = 3 + 3 , xyz = 3- 3 3 ,
a+ x = 6, b + y = 2, c + z = 2 3 T : 3+ 3 3 3 3- 3 3 3 6.2.2 2 3 3
(pcm). 0,50Cu 5a. Gi I, J, K ln lt l trung im ca (2,0)QN, MN, PQ.
Khi : A MB MNBJ =(trung tuyn vung MBN) J 2 QPQ 0,50Tng t DK =.IN 2
K QMIJ = (IJ l tb MNQ). 2 0,50 PNTng t IK =. D PC2V BD BJ + JI + IK
+ KD. Do: 0,50AC AC ACSABCD .BD (BJ+JI + IK+KD) =(MN+NP+PQ+QM) -
pcm. 0,50 2 2 4Cu5b. Chu vi t gic MNPQ l : (1,0)MN + NP + PQ + QM =
2BJ + 2IK +2DK + 2IJ= 2(BJ + JI + IK + KD) 2BD (cmt) 0,50 Du bng xy
ra khi ng gp khc trng vi BD, tc l MQ //NP, MN//PQ, MN=PQ (v cng l
cnh huyn 2 tam gic vung cn bng nhau), lc MNPQ l hnh ch nht.0,50Cu
6. K hiu nh hnh v.(3,0) Phn thun :y AOB =AMB 900(gi thit) t gic
AOBM lun ni tipP HQ x ABO 450 (v AOB AMO M A M vung cn ti O) B Suy
ra M lun nm trn ng thng i qua O v to vi ngO PQ mt gc 45 0.0,50 Trng
hp B v tr B th MB nm trn ng thng i qua O 0,500 v to vi PS mt gc 45
.SKR Gii hn : *) Khi A H th M Q, khi A K th M S 0,50 *) Trng hp B v
tr B: khi A H th M P, khi A K th M R Phn o: Ly M bt k trn ng cho SQ
(hoc M trn PR), qua M k ng thng song song vi ng thng PQ ct (O) ti
A. K bn knh OB 0,50 OA. Ta thy t gic AOBM ni tip (v AMO ABO 450 )
Suy ra : AMB AOB 900 . 0,50 0,50 M AM//PQ , PQ PS MB//PS. Kt lun:Qu
tch giao im M l 2 ng cho ca hnh vung PQRS.www.VNMATH.com31 32.
www.VNMATH.comS Gio dc v o to K thi tuyn sinh lp 10BNH DNGTHPT
Chuyn Hng Vng-------------------- Nm hc 2009-2010Mn thi: Ton
(Chuyn) Thi gian lm bi: 150 pht thi chnh thc (khng k thi gian pht
.) --------------------------------------Cu1: Gii phng trnhx2 x 2 2
x 19 2 x 39Cu 2: Gii h phng trnh 2 x y 3 x y 2 0x y 5 0Cu 3: Cho
a,b R tha: 22 a a 3 b b 3 3 Tnh a+ bCu 4 Cho Phng trnh bc hai , x l
n, tham s m:x2 2 m 1 x 2m 0 1- Chng minh phng trnh lun c hai nghim
phn bit vi mi gi tr m. 2- Gi x1,x2 l hai nghim ca phng trnh . Chng
t M = x1 + x2 - x1x2 khng ph thuc vogi tr ca m .Cu 5 Cho tam gic
ABC c 3 gc nhn . BE v CF l hai ng cao. Trc tm H. Trn HB v HCln lt
ly im M , N sao cho ANB 900 . Chng minh : AM = AN .AMC
--------------------------------www.VNMATH.com32 33.
www.VNMATH.comGiI ThiCu1: Gii phng trnhx2 x 2 2 x 19 2 x 39 (*)t t
= x 2 2 x 19 0(*) t 2 t 2 0 t 4(nhn)1 t2 5(loi x 2 2 x 19 16 x 2 2
x 35 0 x1 7 x 2 5Cu 2: Gii h phng trnh 2 x y 3 x y 2 0(*)x y 5 0t t
= x + y t 1(*) t 2 3t 2 0 1 t2 2 x 3 x y 1 y 2 x y 5 x 7 x y 2 2 x
y 5 3 y 2Cu 3: Cho a,b R tha: 2 2 a a 3 b b 3 3 Tnh a+ b t a a2 3 b
b2 3 3 a2 a2 3 . b 2 b2 3 3 a a2 3 b b2 3 a a2 3 b b2 3 3 2 2 a a 3
b b 3 3 vy a a2 3 b b2 3 3 www.VNMATH.com 33 34. www.VNMATH.com22 2
2ab + a b + 3 + b a + 3 + a + 3 b + 3 = 3 2 2 2 2ab - a b + 3 - b a
+ 3 + a + 3 b + 3 = 3 2a b2 + 3 + 2b a 2 + 3 = 0 a b2 + 3 + b a 2 +
3 = 0 v a 2 + 3 > 0, b2 + 3 > 0 n n a = b = 0 a+b=0Cu 4 Cho
Phng trnh bc hai , x l n, tham s m: x2 2 m 1 x 2m 0 1. =
[-(m+1)]2-2m = m2 +2m +1 -2m = m2 + 1 > 0 Nn phng trnh lun c hai
nghim phn bit vi mi gi tr m 2. TheoViet : x1 + x2 = 2(m + 1) x1.x 2
= 2m M = x1 + x2 - x1.x 2 = 2(m + 1) - 2m = 2 Nn khng ph thuc vo gi
tr ca m .Cu 5:AEB AFC(g-g)A AE AB AF AC AE. AC AF. AB (1) vMAC, ME
: ng .cao E MA2 AE. AC (2) vNAB, NF : ng .cao NA2 AF. AB (3)F H T
(1),(2),(3) N MA2 = NA2 M MA = NA
BC---------------------------------------www.VNMATH.com34 35.
www.VNMATH.comwww.VNMATH.com35 36. www.VNMATH.comwww.VNMATH.com36
37. www.VNMATH.comwww.VNMATH.com37 38.
www.VNMATH.comwww.VNMATH.com38 39. www.VNMATH.comwww.VNMATH.com39
40. www.VNMATH.comHng dnwww.VNMATH.com40 41.
www.VNMATH.comwww.VNMATH.com41 42. www.VNMATH.com Cu
4www.VNMATH.com42 43. www.VNMATH.comwww.VNMATH.com43 44.
www.VNMATH.comwww.VNMATH.com44 45. www.VNMATH.comwww.VNMATH.com45
46. www.VNMATH.comS GIO DC - O TO THI THI TUYN SINH VO LP 10 THPT
CHUYN THI BNHNm hc : 2009-2010BNHMn thi: TON (Dnh cho th sinh thi
vo chuyn Ton, Tin) Thi gian lm bi:150 pht (khng k thi gian giao )
chnh thc thi gm : 01 trang Bi 1. (2,0 im) : a. Cho k l s nguyn dng
bt k. Chng minh bt ng thc sau:11 1 2() ( k 1) k k k 1 1 1 1 188b.
Chng minh rng: 2 3 2 4 3 2010 2009 45 Bi 2. (2.5 im): Cho phng trnh
n x: x 2 ( m 1) x 6 0 (1)(m l tham s)a. Tm cc gi tr ca m phng trnh
(1) c nghim x 1 2b. Tm cc gi tr ca m phng trnh (1) c 2 nghim x1 ,
x2 sao cho biu thc: A ( x12 9)( x2 4) t gi tr ln nht.2 Bi 3. (2,0
im): 22 x y xy 3 a. Gii h phng trnh sau : 33 x y 9 b. Tm cc s nguyn
x, y tha mn phng trnh: x 3 2 x 2 3x 2 y 3 Bi 4. (3,0 im): Cho hnh
vung ABCD tm O, cnh a. M l im di ng trn on OB(M khng trng vi O; B).
V ng trn tm I i qua M v tip xc vi BC ti B, vng trn tm J i qua M v
tip xc vi CD ti D. ng trn (I) v ng trn (J) ct nhau tiim th hai l
N.a. Chng minh rng 5 im A, N, B, C, D cng thuc mt ng trn. T suy ra
3 imC, M, N thng hng.b. Tnh OM theo a tch NA.NB.NC.ND ln nht. Bi 5.
(0.5 im): Cho gc xOy bng 120o , trn tia phn gic Oz ca gc xOy ly im
A sao cho di on thng OA l mt s nguyn ln hn 1. Chng minh rng lun tn
ti t nht bang thng phn bit i qua A v ct hai tia Ox, Oy ln lt ti B v
C sao cho di ccon thng OB v OC u l cc s nguyn dng. ========= Ht
=========Cn b coi thi khng gii thch g thm H v tn th sinh:...S bo
danh:. www.VNMATH.com 46 47. www.VNMATH.comS GIO DC O TO THI K THI
TUYN SINH VO LP 10 THPT CHUYN THI BNHBNHNm hc : 2009-2010HNG DN CHM
V BIU IM MN TON CHUYNCU NI DUNG IMBi 1. a. Cho k l s nguyn dng bt
k. CMR:(2im)11 1 2() (k 1) k k k 1 1 1 1 188 b. Chng minh rng: 2 3
2 4 3 2010 2009 45a. 1 2 k 1 2 k Bt 0.25 (k 1) kk. k 1(1.0) 2k 1 2
k(k 1) 0 0.25 ( k 1 k )2 0 0.25 Lun ng vi mi k nguyn dng. 111
2()0.25 ( k 1) k kk 1b. p dng kt qu cu a ta c:(1.0) 1 1 1 1 0.25 VT
2 1 3 2 4 3 2010 2009 1 1 11 1 1 2 2 20.25 12 2 3 2009 2010 1 2 1
0.25 2010 1 880.25 2 1 VP (pcm) 45 45 www.VNMATH.com47 48.
www.VNMATH.comBi 2 Cho phng trnh n x:x 2 ( m 1) x 6 0 (1) (m
l(2.5tham s)im)c. Tm cc gi tr ca m phng trnh c nghim x 1 2d. Tm m
(1) c 2 nghim x1 , x2 sao cho biu thc:A ( x12 9)( x2 4) max 2 a. 2
(1,5)Pt (1) c nghim x 1 2 1 2 m 1 1 2 6 00.5Tm c m 5 2 6 v KL. 1.0
b. 2Tnh m 1 24 0 m suy ra pt (1) c 2 nghim phn bit (1,0) 0.5x1 , x2
.2 2A x1 x2 6 2 x1 3 x2 2 0.25Theo L Vi-et ta c x1 x2 6 A 2 x1 3 x2
0 2 x1 3 x2 0 x1 3 x1 3 Max A = 0 khi v ch khi x1 x2 6 x2 2 x2 2 x
x 1 m m 0m 20.25 1 2 KL : Vy m = 0 ; m = 2 l cc gi tr cn tm.Bi 3 22
x y xy 3(2 im)a. Gii h phng trnh sau : 3 3 x y 9b. Tm cc s nguyn x,
y tha mn phng trnh:x 3 2 x 2 3x 2 y 3a H phng trnh cho (1.0) x 2 y
2 xy 3x y3 0.52 2 2 ( x y )( x y xy ) 9 ( x y ) 3 xy 3 x y 3 x 1x 2
hoc 0.5 xy 2 y 2y 1b3 72 (1.0) Ta c y 3 x3 2 x 2 3 x 2 2 x 0x y 4 8
0.25(1) 2 33 2 9 15 ( x 2) y 4 x 9 x 6 2 x 0 y x24 160.25(2)T (1) v
(2) ta c x < y < x+2 m x, y nguyn suy ra y = x +
10.25www.VNMATH.com 48 49. www.VNMATH.comThay y = x + 1 vo pt ban u
v gii phng trnh tm c x = -1; x 0.25= 1 t tm c hai cp s (x, y) tha
mn bi ton l (1 ; 2), (-1 ; 0)Bi 4. Cho hnh vung ABCD tm O, cnh a. M
l im di ng trn(3 im)on OB(M khng trng vi O; B). V ng trn tm Ii qua
M v tip xc vi BC ti B, v ng trn tm J i qua M vtip xc vi CD ti D. ng
trn (I) v ng trn (J) ct nhau tiim th hai l N.c. Chng minh rng 5 im
A, N, B, C, D cng thuc mt ng trn.T suy ra 3 im C, M, N thng hng.d.
Tnh OM theo a tch NA.NB.NC.ND ln nht.N AIBKH JMO DCa.MNB MBC ( Cng
chn cung BM) 2.0MND MDC ( Cng chn cung DM)1.5BND MNB MND MBC MDC
90Do 5 im A, B, C, D, M cng thuc mt ng trnSuy ra NC l phn gic ca gc
BND ( do cung BC = cung BD)Mt khc, theo CM trn ta c NM l phn gic ca
gc BND0.5Nn M, N, C thng hng.b. 1.0Gi H, K ln lt l hnh chiu ca N
trn AC v BD0.5 NHOK l hnh ch nhtTa c : NA.NC NH . AC NH .a 2NB.ND
NK .BD NK .a 2www.VNMATH.com raSuy 49 50. www.VNMATH.comNH 2 NK 2
a4NA.NB.NC.ND 2a 2 .NH .NK 2a 2 . a 2 .NO 2 22 a(2 2) a Du bng xy
ra khi v ch khi NH NK OM 0.5 22Bi 5. Cho gc xOy bng 120o , trn tia
phn gic Oz ca gc xOy ly im(0.5 A sao cho di on thng OA l mt s nguyn
ln hn 1. Chngim)minh rng lun tn ti t nht ba ng thng phn bit i qua A
v ct hai tia Ox, Oy ln lt ti B v C sao cho di cc on thng OB v OC u
l cc s nguyn dng. OBAC xz Ch ra ng thng d1 i qua A v vung gc vi OA
tha mn bi ton t OA = a > 1 (a nguyn). Trn tia Ox ly im B sao cho
OB = a + 1 nguyn dng. ng thng d 2 i qua A, B ct tia Oy ti C.111
Chng minh c OB OC OA111 0.5 OC a (a 1) ls nguyna 1 OC a dngSuy ra
d2 l mt ng thng cn tm. Tng t ly B trn Ox sao cho OB = a(a + 1), Ta
tm cng thng d3 Chng minh d1 , d 2 , d3 phn bit. PCMHng dn chung1.
Trn y ch l cc bc gii v khung im cho tng cu. Yu cu hc sinh phi trnh
by, lp lun v bin i hp l, cht ch mi cho im ti a.2. Bi 4 phi c hnh v
ng v ph hp vi li gii bi ton mi cho im.( khng cho im hnh v )3. Nhng
cch gii khc ng vn cho im ti a.www.VNMATH.com50 51. www.VNMATH.com
4. Chm im tng phn, im ton bi l tng cc im thnh phn( khng lm
trn).===========================S GIO DC V O TO K THI TUYN SINH VO
LP 10 CHUYN GIA LAI Nm hc 2009 2010.. CHNH THC.Mn thi: Ton ( Chuyn)
Thi gian lm bi: 150 pht ( Khng k thi gian pht ) BI:Cu 1: ( 1 im)Tm
cc s nguyn dng n sao cho n2 + 1 chia ht cho n + 1Cu 2: ( 1,5 im) 2
x 9 2 x 1 x 3Cho biu thc A = x 5 x 6 3 xx 2 a) Rt gn A. b) Tm cc gi
tr nguyn ca x A nhn gi tr nguyn.Cu 3: ( 1,5 im) Gi s x1, x2 l hai
nghim ca phng trnh: x2 4x + 1 = 0. Tnh x12 + x22, x13 + x23 v x15 +
x25( khng s dng my tnh cm tay tnh).Cu 4: ( 2 im)a) V th ca cc hm s
y x 1 v y x 2 trn cng mt h trc ta Oxy. b) Chng t phng trnh x 1 x 2
c mt nghim duy nht.Cu 5: ( 1,5 im)Mt ngi d nh ro xung quanh mt ming
t hnh ch nht c din tch 1.600m2, di haicnh l x mt v y mt. Hai cnh k
nhau ro bng gch, cn hai cnh kia ro bng . Mi mt robng gch gi 200.000
ng, mi mt ro bng gi 500.000 ng. a) Tnh gi tin d nh ro ( theo x v
y). b) Ngi y c 55 triu ng, hi s tin y c ro khng ?Cu 6: ( 2,5 im)
Cho tam gic ABC c ba gc nhn ni tip (O;R). Cc ng cao AD, BE, CF ct
nhau ti H.AO ko di ct (O) ti M.a) Chng minh t gic AEHF l t gic ni
tip v t gic BHCM l hnh bnh hnh.b) Chng minh AO EF.www.VNMATH.com 51
52. www.VNMATH.com R2 p 2 c) Chng minh rng: S ABC , trong SABC l
din tch tam gic ABC v p l chu vi4ca tam gic DEF.Ht.H v tn: ...;
SBD.; Phng thi s:......Ch k ca gim th 1:; Ch k ca gim th 2:...S GIO
DC V O TO K THI TUN SINH LP 10 NM HC 2009-2010 LONG AN Mn thi : TON
h chuyn Ngy thi : 10-7 2009 CHNH THCThi gian : 150 pht ( khng k pht
)Cu 1 (2)Rt gn cc biu thc sau :1) A = 4+2 3+ 4-2 3 33 2) B =7+5 2+
7-5 2Cu 2 (2) x2x1 + y y 1 = 6- - 1) Gii h phng trnh : x 3y x - 1 +
y - 1 = 82) Gii phng trnh : x4 - 2x3 - x2 + 2x + 1 = 0Cu 3 (2) Gi
th hm s y = x2 l parabol (P), th ca hm s y = x - m l ng thng (d) .
1) Tm gi tr ca m (d) ct (P) ti hai im phn bit . 2) Khi (d) ct (P)
ti hai im phn bit A v B k hiu xA v xB ln lt l honh ca A v B . Tm cc
gi tr ca m sao cho x3 + x3 = 1 .A BCu 4 (2)1) Cho tam gic ABC . Gi
M,N,P ln lt l trung im ca cc cnh AB,BC,CA.Khng nh SABC = 4SMNP ng
hay sai ? ti sao ?2) Cho ng trn (T) c ng knh AB . Gi C l im i xng
vi A qua B , PQ l mtng knh thay i ca (T) khc ng knh AB. ng thng CQ
ct ng thng PB im M . Khng nh CQ = 2CM ng hay sai ? ti sao ?Cu 5 (2)
1) Cho hai s thc x , y thay i v tho mn iu kin : 2x + 3y = 5 . Tm x
,y biu thc P = 2x2 + 3y2 + 2 t gi tr nh nht . Tm gi tr nh nht . 2)
Cho t , y l hai s thc tho mn iu kin : t + y2 + y t - 5 t - 4y + 7 =
0. Hy tm t , y .Ht S gio dc v o toHng yn chnh thcwww.VNMATH.com 52
53. www.VNMATH.comk thi tuyn sinh vo lp 10 thpt chuyn Nm hc 2009
2010 Mn thi: Ton (Dnh cho th sinh thi vo cc lp chuyn Ton, Tin) Thi
gian lm bi: 150 phtBi 1: (1,5 im) 1 1 Cho a 2 : 7 1 1 7 1 1 Hy lp
mt phng trnh bc hai c h s nguyn nhn a - 1 l mt nghim.Bi 2: (2,5
im)x 16 xy y 3 a) Gii h phng trnh: xy y 9x 22 b) Tm m phng trnh x 2
2x 3x 2 6x m 0 c 4 nghim phn bit.Bi 3: (2,0 im) a) Chng minh rng nu
s nguyn k ln hn 1 tho mn k 2 4 v k 2 16 l cc snguyn t th k chia ht
cho 5. b) Chng minh rng nu a, b, c l di ba cnh ca mt tam gic c p l
na chu vi thp a p b p c 3pBi 4: (3,0 im) Cho ng trn tm O v dy AB
khng i qua O. Gi M l im chnh gia ca cung ABnh. D l mt im thay i trn
cung AB ln (D khc A v B). DM ct AB ti C. Chng minh rng: a) MB.BD
MD.BC b) MB l tip tuyn ca ng trn ngoi tip tam gic BCD. c) Tng bn
knh cc ng trn ngoi tip tam gic BCD v ACD khng i.Bi 5: (1,0 im) Cho
hnh ch nht ABCD. Ly E, F thuc cnh AB; G, H thuc cnh BC; I, J thuc
cnh CD;K, M thuc cnh DA sao cho hnh 8 - gic EFGHIJKM c cc gc bng
nhau. Chng minh rng nu di cc cnh ca hnh 8 - gic EFGHIJKM l cc s hu
t th EF = IJ.------------ Ht ------------H v tn thS bo
danh:......Phng thi s:......sinh:.........Ch k ca gim
th.....................www.VNMATH.com 53 54. www.VNMATH.com Hng dn
chm thiBi 1: (1,5 im) 117 1 17 1 1 a 2: 2:0,5 7 1 17 1 17 2a= 2: 7
0,25 7t x a 1 x 7 1 x 1 7 x 2 2x 1 7 0,5 x 2 2x 6 0 0,25 Vy phng
trnh x 2 2x 6 0 nhn7 1 lm nghimBi 2: (2,5 im) x 16x 16 xy xy (1) y
3 y 3a) K: x, y 00,25 xy y 9 y x 5(2) x 2 x y 6Gii (2) 6y 2 6x 2
5xy (2x 3y)(3x 2y) 00,25 3y* Nu 2x 3y 0 x .2 0,25 3y 3 16Thay vo
(1) ta cy. 2 2 3 3y2 23 (phng trnh v nghim) 0,25 2 62y* Nu 3x 2y 0
x . 3 0,25 Thay vo (1) ta c y 2 9 y 3- Vi y 3 x 2 (tho mn iu kin)-
Vi y 3 x 2 (tho mn iu kin) 0,25 Vy h phng trnh c hai nghim: (x; y)
= (2; 3); (x; y) = (-2; -3) 2b) t x 2 2x 1 y x 1 y x 1 y(y 0)
(*)2Phng trnh cho tr thnh: y 1 3 y 1 m 0 0,25 y 2 5y m 4 0 (1)T (*)
ta thy, phng trnh cho c 4 nghim phn bit th phng trnh (1) c 0,25 2
nghim dng phn bit 0 9 4m 0 S 0 5 00,25 P 0 m 4 0www.VNMATH.com54
55. www.VNMATH.com 9 m 9 4 4 m m 4 4 0,25 9Vy vi 4 m th phng trnh c
4 nghim phn bit.4Bi 3: (2,0 im)a) V k > 1 suy ra k 2 4 5; k 2 16
5- Xt k 5n 1 (vi n ) k 2 25n 2 10n 1 k 2 4 5 0,25 k 2 4 khng l s
nguyn t.- Xt k 5n 2 (vi n ) k 2 25n 2 20n 4 k 2 16 5 0,25 k 2 16
khng l s nguyn t.- Xt k 5n 3 (vi n ) k 2 25n 2 30n 9 k 2 16 5 0,25
k 2 16 khng l s nguyn t.- Xt k 5n 4 (vi n ) k 2 25n 2 40n 16 k 2 4
5 k 2 4 khng l s nguyn t. 0,25 Do vy k 5 2 b) Ta chng minh: Vi a,
b, c th a b c 3 a 2 b 2 c 2 (*) Tht vy (*) a b c 2ab 2bc 2ca 3a 3b
3c 22 2 22 2 0,5 (a b)2 (b c)2 (c a)2 0 (lun ng)p dng (*) ta c:2 pa
pb pc 3 3p a b c 3p0,5 Suy ra p a p b p c 3p (pcm)Bi 4: (3,0 im) N
DJ I OA C B Mwww.VNMATH.com55 56. www.VNMATH.coma) Xt MBC v MDB c:
BDM MBC (hai gc ni tip chn hai cung bng nhau) 0,5 BMC BMDDo vy MBC
v MDB ng dngMB MD0,5 Suy ra MB.BD MD.BCBC BD b) Gi (J) l ng trn
ngoi tip BDC BJC 2BDC 2MBCBJChay MBC 2 0,5 0 180 BJCBCJ cn ti J CBJ
2 O BJC 180 BJC 90 O MB BJSuy ra MBC CBJ 2 20,5 Suy ra MB l tip
tuyn ca ng trn (J), suy ra J thuc NBc) K ng knh MN ca (O) NB MBM MB
l tip tuyn ca ng trn (J), suy ra J thuc NBGi (I) l ng trn ngoi tip
ADC 0,5 Chng minh tng t I thuc AN Ta c ANB ADB 2BDM BJC CJ //
INChng minh tng t: CI // JNDo t gic CINJ l hnh bnh hnh CI = NJSuy
ra tng bn knh ca hai ng trn (I) v (J) l: 0,5 IC + JB = BN (khng
i)Bi 5: (1,0 im)A E a F BbGh cM HgK df eD J I CGi EF = a ; FG = b ;
GH = c ; HI = d ; IJ = e ; JK = f ; KM = g ; ME = h (vi a, b,c, d,
e, f, g, h l cc s hu t dng) 0,25 Do cc gc ca hnh 8 cnh bng nhau nn
mi gc trong ca hnh 8 cnh c s o(8 2).180Ol: 135O8Suy ra mi gc ngoi
ca hnh 8 cnh l: 180O - 135O = 45 O 0,5 www.VNMATH.com56 57.
www.VNMATH.comDo cc tam gic MAE ; FBG ; CIH ; DKJ l cc tam gic vung
cn. h b df MA = AE = ; BF = BG = ; CH = CI =; DK = DJ =2 2 22h
bfdTa c AB = CD nn: a e 2 222 (e - a) 2 = h + b - f - d h bf dNu e
- a 0 th 2 (iu ny v l do 2 l s v t)ea 0,25 Vy e - a = 0 e = a hay
EF = IJ (pcm).------------ Ht ------------www.VNMATH.com57 58.
www.VNMATH.comS gio dc v o to K thi tuyn sinh lp 10 THPT chuynHI
dng nguyn tri - Nm hc 2009-2010 Mn thi : tonThi gian lm bi: 150 pht
thi chnh thcNgy thi 08 thng 7 nm 2009( thi gm: 01 trang)Cu I (2.5
im): 1) Gii h phng trnh:x 2 y 2 xy 3 2 xy 3x 4 2) Tm m nguyn phng
trnh sau c t nht mt nghim nguyn: 4x 2 4mx 2m 2 5m 6 0Cu II (2.5
im): 1) Rt gn biu thc: 2 4 x2 2 x 33 2 x A vi 2 x 2 4 4 x23 2) Cho
trc s hu t m sao cho m l s v t. Tm cc s hu t a, b, c : 3 2a m b3 m
c 0Cu III (2.0 im): 1) Cho a thc bc ba f(x) vi h s ca x 3 l mt s
nguyn dng v bitf(5) f(3) 2010 . Chng minh rng: f(7) f(1) l hp s. 2)
Tm gi tr ln nht ca biu thc: P x 2 4x 5 x 2 6x 13Cu IV (2.0 im): Cho
tam gic MNP c ba gc nhn v cc im A, B, C ln lt l hnh chiu vung gc
caM, N, P trn NP, MP, MN. Trn cc on thng AC, AB ln lt ly D, E sao
cho DE song song vi NP. Trn tia AB ly im K sao cho DMK NMP . Chng
minh rng: 1) MD = ME 2) T gic MDEK ni tip. T suy ra im M l tm ca ng
trn bng tip gc DAKca tam gic DAK.Cu V (1.0 im): Trn ng trn (O) ly
hai im c nh A v C phn bit. Tm v tr ca cc im B v Dthuc ng trn chu vi
t gic ABCD c gi tr ln
nht.-----------------------Ht-----------------------H v tn th sinh
: ......................................................S bo danh
:.......................Ch k ca gim th 1 :
.............................Ch k ca gim th
2:............................ Hng dn chmwww.VNMATH.com58 59.
www.VNMATH.com Cu Phn ni dung im2 2cu I1)x y xy 3 (1) 22,5 im 1,5im
xy 3x 4 (2) 4 3x 2 T (2) x 0. T y , thay vo (1) ta c:x0.25 222 4 3x
4 3x x2 x.3 x x 0.25 7x 4 23x2 16 00.25 16 Gii ra ta c x 2 1 hoc x2
=70.25164 75 7 T x 2 1 x 1 y 1 ; x 2 x y 7 770.25 4 7 5 7 4 7 5 7 7
; 7 7 ; 7 Vy h c nghim (x; y) l (1; 1); (-1; -1); ; 0.25 2)iu kin
phng trnh c nghim: x 00.25 1,0im m 5m 6 0 (m 2)(m 3) 0 . V (m - 2)
> (m - 3) nn: x 0 m 2 0 v m 3 0 2 m 3, m m Z m = 2 hoc m = 3.
0.25Khi m = 2 x= 0 x = -1 (tha mn)Khi m = 3 x= 0 x = - 1,5
(loi).0.25 Vy m = 2.0.25cu II 1)t a 2 x; b 2 x(a, b 0)2 22 22,5 im
1,5im a b 4; a b 2x 0.252 ab a 3 b3 2 ab a b a 2 b 2 ab A 4 ab 4
ab0.252 ab a b 4 ab A 2 ab a b 4 ab0.25 A 2 4 2ab a b 0.25 A 2 a 2
b2 2ab a b a b a b 0.25 A 2 a 2 b2 2x A x 2 0.25 2)a 3 m 2 b 3 m c
0 (1) 1,0im Gi s c (1) b 3 m 2 c 3 m am 0 (2) T (1), (2) (b2 ac) 3
m (a 2 m bc)0.25 www.VNMATH.com 59 60. www.VNMATH.com 2a 2 m bc3 Nu
a m bc 0 m 2 l s hu t. Tri vi gi thit!b ac b2 ac 0 b3 abc 22 a m bc
0 bc am 0.25 b b3 a 3m b a 3 m . Nu b 0 th 3 m l s hu t. Tri vi gi
thit! a a 0;b 0 . T ta tm c c = 0. 0.25 Ngc li nu a = b = c = 0 th
(1) lun ng. Vy: a = b = c = 00.253 2cu III1)Theo bi ra f(x) c dng:
f(x) = ax + bx + cx + d vi a nguyn dng. 0.252 im1,0imTa c: 2010 =
f(5) - f(3) = (53 - 33)a + (52 - 32)b + (5 - 3)c = 98a + 16b + 2c
16b + 2c = (2010- 98a) 0.25 Ta c f(7) - f(1) = (73 - 13)a + (72 -
12)b + (7 - 1)c = 342a + 48b + 6c = 342a + 3(16b + 2c) = 342a +
3(2010- 98a)= 48a + 6030 = 3.(16a + 2010) 3 0.25 V a nguyn dng nn
16a + 2010>1 . Vy f(7)-f(1) l hp s0.25 2)2 2 Px 2 12 x 3 221,0im
Trn mt phng ta Oxy ly cc im A(x-2; 1), B(x+3; 2) 0.2522 Ta chng
minh c: AB x 2 x 3 1 2 25 1 2622OA x 2 12 , OB x 3 220.252 2 Mt khc
ta c: OA OB AB x 2 12 x 3 2 2 26 0.25 Du = xy ra khi A thuc on OB
hoc B thuc on OAx2 1 x 7 .Th li x = 7 th A(5; 1); B(10; 2) nn A
thuc onx3 2 OB. Vy Max P 26 khi x = 7. 0.25cuIV1) Ta d dng chng
minh t gic MBAN ni tip MAB MNB ,2 im0,75im M MCAP ni tip CAM CPM
.0.25 Li c BNM CPM K(cng ph gc NMP)B CAM BAM (1)0.25 CDo DE // NP
mt khc D EMA NP MA DE (2)T (1), (2) ADE cn ti A MA l trung trc ca
DE MD = ME NP0.25 A 2)1,25im 0.25www.VNMATH.com 60 61.
www.VNMATH.comM KBCD E NPA Do DE//NP nn DEK NAB , mt khc t gic MNAB
ni tip nn: NMB NAB 1800 NMB DEK 1800 Theo gi thit DMK NMP DMK DEK
1800 T gic MDEK ni tip0.25 Do MA l trung trc ca DE MEA MDA 0.25 MEA
MDA MEK MDC . 0.25 V MEK MDK MDK MDC DM l phn gic ca gc CDK, kt hp
vi AM l phn gic DAB M l tm ca ng trn bng tip gc DAK ca tam gic
DAK.0.25cu VA1 imBBOA CD D Khng mt tng qut gi s:AB AC. Gi B l im
chnh gia cung ABC AB CBTrn tia i ca BC ly im A sao cho BA = BA AB
BC CA 0.25 Ta c: B BC B AC B CA (1) ; B CA B BA 1800 (2) B BA 1800
(3);T (1), (2), (3) B BA B BAB BC 0.25 Hai tam gic ABB v ABB bng
nhau AB BA Ta c B A BC BA BC A C = AB + BC ( BA + BC khng i v B, A,
C c nh). Du = xy ra khi B trng vi B. 0.25 Hon ton tng t nu gi D l
im chnh gia cung ADC th ta cng c AD + CD AD + CD. Du = xy ra khi D
trng vi D. Chu vi t gic ABCD ln nht khi B, D l cc im chnh gia cc
cung AC ca ng trn (O)0.25www.VNMATH.com 61 62. www.VNMATH.comCh :
Nu th sinh lm theo cch khc, li gii ng vn cho im ti a.www.VNMATH.com
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