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B n thi vo THPT Nm hc 2009 - 2010
Su tm: ON TIN TRUNG - THCS Hong Vn Th - N 1
1Bi 1 : (2 im)a) Tnh :
b) Gii h phng trnh :
Bi 2 : (2 im)Cho biu thc :
a) Rt gn A.b) Tm x nguyn A nhn gi tr nguyn.
Bi 3 : (2 im)Mt ca n xui dng t bn sng A n bn sng B cch nhau 24 km ; cng lc, cng t A v B mt b na tri vi vn tc dng nc l 4 km/h. Khi n Bca n quay li ngay v gp b na ti a im C cch A l 8 km. Tnh vn tcthc ca ca n.Bi 4 : (3 im)Cho ng trn tm O bn knh R, hai im C v D thuc ng trn, B ltrung im ca cung nh CD. Kng knh BA ; trn tia i ca tia AB lyim S, ni S vi C ct (O) ti M ; MD ct AB ti K ; MB ct AC ti H.a) Chng minh BMD = BAC, t => t gic AMHK ni tip.
b) Chng minh : HK // CD.c) Chng minh : OK.OS = R2.Bi 5 : (1 im)Cho hai s a v b khc 0 tha mn : 1/a + 1/b = 1/2Chng minh phng trnh n x sau lun c nghim : (x2 + ax + b)(x2 + bx + a)= 0.
Hng dn gii
Bi 3:Do ca n xut pht t A cng vi b na nn thi gian ca ca n bng thi gian b na:
8 24
= (h)
Gi vn tc ca ca n l x (km/h) (x>4)
Theo bi ta c:24 24 8 24 16
2 24 4 4 4 x x x x
+ = + =
+ +
2 02 40 020
xx x
x
= =
=
Vy vn tc thc ca ca n l 20 km/h
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Bi 4:a) Ta c BC BD= (GT) BMD BAC= (2 gcni tip chn 2 cung bng nhau)* Do BMD BAC= A, M nhn HK di 1 gc
bng nhau MHKA ni tip.b) Do BC = BD (do BC BD= ), OC = OD (bnknh) OB l ng trung trc ca CD CD AB (1)Xet MHKA: l t gic ni tip, 090AMH= (gcnt chn na ng trn) 0 0 0180 90 90HKA = = (l) HK AB (2)T 1,2 HK // CD
H K
M A
B
O
C D
S Bi 5:
22 2
2
0 (*)( )( ) 0
0 (**)
x ax b x ax b x bx a
x bx a
+ + =+ + + + =
+ + =
(*) 4b2 = , PT c nghim 2 21 1
4 0 42
a b a ba b
(3)
(**) 2 4b a = PT c nghim th 21 1
4 02
b ab a
(4)
Cng 3 vi 4 ta c:1 1 1 1
2 2a b a b
+ +
1 1 1 1 1 1 1 1 1 1 1 1
2 4 4 4 4 4 8 42 2 a b a ba b
+ + +
(lun lun ng vi mi a, b)
De 2
thi gm c hai trang.
PHN 1. TRC NGHIM KHCH QUAN : (4 im)
1. Tam gic ABC vung ti A c3
tg4
B = . Gi tr cosCbng :
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a).3
cos5
C= ; b).4
cos5
C= ; c).5
cos3
C= ; d).5
cos4
C=
2. Cho mt hnh lp phng c din tch ton phn S1 ; th tch V1 v mt hnh cu c
din tch S2 ; th tch V2. Nu S1 = S2 th t s th tch 12
V
Vbng :
a). 12
V 6
V = ; b). 1
2
V
V 6
= ; c). 1
2
V 4
V 3= ; d). 1
2
V 3
V 4
=
3. ng thc 4 2 28 16 4 x x x + = xy ra khi v ch khi :a).x 2 ; b).x 2 ; c).x 2 vx 2 ; d).x 2 hocx 2
4. Cho hai phng trnhx2 2x + a = 0 v x2 +x + 2a = 0. hai phng trnh cngv nghim th :
a). a > 1 ; b). a < 1 ; c).
1
8a>
; d).
1
8a
>
>
2 2
2
( 4 ) 4(7 1) 0
4 0
7 1 0
m m m
m m
m
+ >
+ >
>
(I) +
Vi iu kin (I), (1) c 2 nghim phn bit dng X1 , X2. phng trnh cho c 4 nghimx1, 2 = 1X ;x3, 4 = 2X
2 2 2 2 21 2 3 4 1 22( ) 2( 4 )x x x x X X m m + + + = + = + +
Vy ta c 2 2 12( 4 ) 10 4 5 05
mm m m mm
=+ = + = =
+
Vi m = 1, (I) c tha mn +Vi m = 5, (I) khng tha mn. +Vy m = 1.
2.t 4 2 1t x x= + + (t 1)
c phng trnh3
5 3( 1)tt
+ = +
3t2 8t 3 = 0
t = 3 ;1
3t= (loi) +
Vy 4 2 1 3x x+ + = x = 1. +
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Cu 2 : (3,5 im)1.
2 2 2 2cos 2 1 sin 1 cos 2 cos 1P = + = + 2cos 2cos 1P = + (v cos > 0) +
2(cos 1)P = +
1 cosP = (v cos < 1) +
2.
( )( ) ( ) ( ) ( )2
4 15 5 3 4 15 5 3 4 15 4 15+ = + +
= ( )5 3 4 15 +
= ( ) ( )2
5 3 4 15 + +
=( )( )8 2 15 4 15 + +
= 2 +
Cu 3 : (2 im)
( )2
0 2a b a b ab + +
Tng t, 2a c ac+ 2b c bc+
1 2a a+ +1 2b b+ 1 2c c+
Cng v vi v cc bt ng thc cng chiu trn ta c iu phi chng minh.+
ng thc xy ra a = b = c = 1+
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Cu 4 : (6 im)
+
1.Ta c : ABC = 1v
ABF = 1v
B, C, F thng hng. +AB, CE v DF l 3 ng cao ca tam gic ACF nn chng ng quy. ++
2.ECA = EBA (cng chn cung AE ca (O) +M ECA = AFD (cng ph vi hai gc i nh) + EBA = AFD hay EBI = EFI + T gic BEIF ni tip. +
3.
Gi H l giao
im ca AB v PQChng minh c cc tam gic AHP v PHB ng dng +
HP HA
HB HP= HP2 = HA.HB +
Tng t, HQ2 = HA.HB + HP = HQ H l trung im PQ. +
Lu :- Mi du + tng ng vi 0,5 im.- Cc cch gii khc c hng im ti a ca phn .
- im tng phn, im ton bi khng lm trn.
3I.Trc nghim:(2 im)
O O
B
A
C
D
E
F
I
P
QH
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Hy ghi li mt ch ci ng trc khng nh ng nht.
Cu 1: Kt qu ca php tnh ( )8 18 2 98 72 : 2 + l :
A . 4 B . 5 2 6+ C . 16 D . 44
Cu 2 : Gi tr no ca m th phng trnh mx2 +2 x + 1 = 0 c hai nghim phn
bit :
A. 0m B. 14
m < C. 0m v 14
m < D. 0m v 1m <
Cu 3 :Cho ABC ni tip ng trn (O) c 0 060 ; 45B C= = . SBC l:
A . 750 B . 1050 C . 1350 D . 1500
Cu 4 : Mt hnh nn c bn knh ng trn y l 3cm, chiu cao l 4cm th
din tch xung quanh hnh nn l:
A 9(cm2) B. 12(cm2) C . 15(cm2) D. 18(cm2)
II. T Lun: (8 im)
Cu 5 : Cho biu thc A= 1 21 1
x x x x
x x
+ ++
+
a) Tm x biu thc A c ngha.
b) Rt gn biu thc A.
c) Vi gi tr no ca x th ABC). V ng trn tm (O') ng knh BC.Gi I l trung im
ca AC. V dy MN vung gc vi AC ti I, MC ct ng trn tm O' ti
D.a) T gic AMCN l hnh g? Ti sao?
b) Chng minh t gic NIDC ni tip?
c) Xc nh v tr tng i ca ID v ng trn tm (O) vi ng trn
tm (O').
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p n
Cu Ni dung im1 C 0.52 D 0.5
3 D 0.54 C 0.55
a) A c ngha 0
1 0
x
x
0
1
x
x
0.5
b) A=( ) ( )
2
1 1
1 1
x x x
x x
+
+ +
0.5
= 1x x + 0.25
=2 1x 0.25
c) A
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I
D
N
M
O'O
A
C
B
a) ng knh AB MN (gt) I l trung im ca MN (ngknh v dy cung)
0.5
IA=IC (gt) T gic AMCN c ng cho AC v MN ct nhau titrung im ca mi ng v vung gc vi nhau nn l hnh thoi.
0.5
b) 090ANB = (gc ni tip chn 1/2 ng trn tm (O) ) BN AN.AN// MC (cnh i hnh thoi AMCN).
BN MC (1)090BDC= (gc ni tip chn 1/2 ng trn tm (O') )
BD MC (2)T (1) v (2) N,B,D thng hng do 090NDC= (3).
090NIC= (v AC MN) (4)
0.5
T (3) v (4) N,I,D,C cng nm trn ng trn ng knh NC T gic NIDC ni tip 0.5c) OBA. O'BC m BA vafBC l hai tia i nhau B nm gia Ov O' do ta c OO'=OB + O'B ng trn (O) v ng trn
(O
'
) tip xc ngoi ti B
0.5
MDN vung ti D nn trung tuyn DI = 12
MN =MI MDI cn
IMD IDM= .Tng t ta c ' 'O DC O CD= m 0' 90 IMD O CD+ = (v 090MIC= )
0.25 0' 90 IDM O DC+ = m 0180MDC= 0' 90IDO = do ID DO ID l tip tuyn ca ng trn (O'). 0.25
Ch : Nu th sinh lm cch khc ng vn cho im ti a
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4Cu1 : Cho biu thc
A=2
)1(:
1
1
1
12
2233
+
+
+
x
xxx
x
xx
x
xVi x 2 ;1
.a, Ru gn biu thc A
.b , Tnh gi tr ca biu thc khi cho x= 226+
c. Tm gi tr ca x A=3Cu2.a, Gii h phng trnh:
=+
=+
1232
4)(3)( 2
yx
yxyx
b. Gii bt phng trnh:
3
15242
23
++
xx
xxx x =2
173
Cu 2 : a)t x - y= a ta c pt: a2+3a=4 => a=-1; a=-4
T ta c
=+
=+
1232
4)(3)( 2
yx
yxyx
*
=+
=
1232
1
yx
yx(1)
*
=+
=
1232
4
yx
yx(2)
Gii h (1) ta c x=3, y=2Gii h (2) ta c x=0, y=4Vy h phng trnh c nghim l x=3, y=2 hoc x=0; y=4
b) Ta c x3- 4x2- 2x- 15 = (x-5)(x2+x+3)m x2+x+3=(x+1/2)2+11/4>0 vi mi xVy bt phng trnh tng ng vi x-5>0 =>x>5Cu 3: Phng trnh: ( 2m-1)x2-2mx+1=0 Xt 2m-1=0=> m=1/2 pt tr thnh x+1=0=> x=1
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O
K
F
E
D
B
A
Xt 2m-10=> m 1/2 khi ta c,
= m2-2m+1= (m-1)20 mi m=> pt c nghim vi mi mta thy nghim x=1 khng thuc (-1,0)
vi m 1/2 pt cn c nghim x=12
1
+
m
mm=
12
1
m
pt c nghim trong khong (-1,0)=> -1m E,F thuc ng trn ng knh BK
hay 4 im E,F,B,K thuc ng trn ng knh BK.b. BCF= BAFM BAF= BAE=450=> BCF= 450Ta c BKF= BEFM BEF= BEA=450(EA l ng cho ca hnh vung ABED)=> BKF=450V BKC= BCK= 450=> tam gic BCK vung cn ti B
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5
Bi 1: Cho biu thc: P =( )
+
+
+
1
122:
11
x
xx
xx
xx
xx
xx
a,Rt gn P
b,Tm x nguyn P c gi tr nguyn.
Bi 2: Cho phng trnh: x2-( 2m + 1)x + m2 + m - 6= 0 (*)
a.Tm m phng trnh (*) c 2 nghim m.
b.Tm m phng trnh (*) c 2 nghim x1; x2 tho mn3
23
1 xx =50
Bi 3: Cho phng trnh: ax2 + bx + c = 0 c hai nghim dng phn bit x 1, x2Chng
minh:
a,Phng trnh ct2 + bt + a =0 cng c hai nghim dng phn bit t1 v t2.
b,Chng minh: x1 + x2 + t1 + t2 4Bi 4: Cho tam gic c cc gc nhn ABC ni tip ng trn tm O . H l trc tm
ca tam gic. D l mt im trn cung BC khng cha im A.
a, Xc nh v tr ca im D t gic BHCD l hnh bnh hnh.
b, Gi P v Q ln lt l cc im i xng ca im D qua cc ng thng AB
v AC . Chng minh rng 3 im P; H; Q thng hng.
c, Tm v tr ca im D PQ c di ln nht.
Bi 5: Cho hai s dng x; y tho mn: x + y 1
Tm gi tr nh nht ca: A =xyyx
501122
++
p n
Bi 1: (2 im). K: x 1;0 x
a, Rt gn: P =( )
( )
( )1
12:
1
122
x
x
xx
xx z P =
1
1
)1(
12
+=
x
x
x
x
b. P =1
21
1
1
+=
+
xx
x
P nguyn th
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)(121
9321
0011
4211
Loaixx
xxx
xxx
xxx
==
===
===
===
Vy vi x= { }9;4;0 th P c gi tr nguyn.
Bi 2: phng trnh c hai nghim m th:
( ) ( )
+=
++=
012
06
06412
21
221
22
mxx
mmxx
mmm
3
2
1
0)3)(2(
025
+
>=
m
m
mm
b. Gii phng trnh: ( ) 50)3(2 33 =+ mm
=
+=
=+=++
2
51
2
51
0150)733(5
2
1
22
m
m
mmmm
Bi 3: a. V x1 l nghim ca phng trnh: ax2 + bx + c = 0 nn ax1
2 + bx1 + c =0. .
V x1> 0 => c. .01
.1
1
2
1=++
a
xb
xChng t
1
1
xl mt nghim dng ca phng
trnh: ct2 + bt + a = 0; t1 =1
1
xV x2 l nghim ca phng trnh:
ax2 + bx + c = 0 => ax22 + bx2 + c =0
v x2> 0 nn c. 01
.1
2
2
2
=+
+
a
xb
xiu ny chng t
2
1
xl mt nghim dng ca
phng trnh ct2 + bt + a = 0 ; t2 =2
1
x
Vy nu phng trnh: ax2 + bx + c =0 c hai nghim dng phn bit x1; x2 th
phng trnh : ct2 + bt + a =0 cng c hai nghim dng phn bit t1 ; t2 . t1 =1
1
x; t2
=2
1
x
b. Do x1; x1; t1; t2 u l nhng nghim dng nn
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t1+ x1 =1
1
x+ x1 2 t2 + x2 =
2
1
x+ x2 2
Do x1 + x2 + t1 + t2 4
Bi 4
a. Gi s tm c im D trn cung BC sao cho t gic BHCD l hnh bnh hnh .Khi : BD//HC; CD//HB v H l trc tm tam gic ABC nn
CH AB v BH AC => BD AB v CD AC .
Do : ABD = 900 v ACD = 900 .
Vy AD l ng knh ca ng trn tm O
Ngc li nu D l u ng knh AD
ca ng trn tm O th
t gic BHCD l hnh bnh hnh.b)V P i xng vi D qua AB nn APB = ADB
nhng ADB = ACB nhng ADB = ACB
Do : APB = ACB Mt khc:
AHB + ACB = 1800 => APB + AHB = 1800
T gic APBH ni tip c ng trn nn PAB = PHB
M PAB = DAB do : PHB = DAB
Chng minh tng t ta c:
CHQ =
DACVy PHQ = PHB + BHC + CHQ = BAC + BHC = 1800
Ba im P; H; Q thng hng
c). Ta thy APQ l tam gic cn nh A
C AP = AQ = AD v PAQ = 2BAC khng i nn cnh y PQ
t gi tr ln nht AP v AQ l ln nht hay AD l ln nht D l u ng knh k t A ca ng trn tm O
HO
P
Q
D
CB
A
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6
Bi 1: Cho biu thc:( ) ( )( )yx
xy
xyx
y
yyx
xP
+
++
+=
111))1)((
a). Tm iu kin ca x v y P xc nh . Rt gn P.b). Tm x,y nguyn tha mn phng trnh P = 2.
Bi 2: Cho parabol (P) : y = -x2 v ng thng (d) c h s gc m i qua im
M(-1 ; -2) .a). Chng minh rng vi mi gi tr ca m (d) lun ct (P) ti hai im A , B
phn bitb). Xc nh m A,B nm v hai pha ca trc tung.
Bi 3: Gii h phng trnh :
=++
=++
=++
27
1111
9
zxyzxy
zyx
zyx
Bi 4: Cho ng trn (O) ng knh AB = 2R v C l mt im thuc ng trn);( BCAC . Trn na mt phng b AB c cha im C , k tia Ax tip xc vi
ng trn (O), gi M l im chnh gia ca cung nh AC . Tia BC ct Ax ti Q , tiaAM ct BC ti N.a). Chng minh cc tam gic BAN v MCN cn .b). Khi MB = MQ , tnh BC theo R.
Bi 5: Cho Rzyx ,, tha mn :zyxzyx ++
=++1111
Hy tnh gi tr ca biu thc : M =4
3+ (x8 y8)(y9 + z9)(z10 x10) .
p n
Bi 1: a). iu kin P xc nh l :; 0;1;0;0 + yxyyx .*). Rt gn
P:( )
( )( )( )
(1 ) (1 )
1 1
x x y y xy x yP
x y x y
+ +
=
+ +
( ) ( )( )( )( )
( )
1 1
x y x x y y xy x y
x y x y
+ + +
=
+ +
( )( )( )( )( )1 1
x y x y x xy y xy
x y x y
+ + +
=
+ +
( ) ( ) ( )( )( ) ( )
1 1 1 1
1 1
x x y x y x x
x y
+ + + +
=
+
( )1 x y y y x
y
+ =
( )( ) ( )( )
1 1 1
1
x y y y y
y
+
=
. x xy y= +
Vy P = .yxyx +
b). P = 2 .yxyx + = 2
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Q
N
M
O
C
BA
( (
( )( ) 111111
=+
=++
yx
yyx
Ta c: 1 + 1y 1 1x 0 4x x = 0; 1; 2; 3 ; 4
Thay vo ta c cc cp gi tr (4; 0) v (2 ; 2) tho mnBi 2: a). ng thng (d) c h s gc m v i qua im M(-1 ; -2) . Nn phngtrnh ng thng (d) l : y = mx + m 2.Honh giao im ca (d) v (P) l nghim ca phng trnh:
- x2 = mx + m 2 x2 + mx + m 2 = 0 (*)
V phng trnh (*) c ( ) mmmm >+=+= 04284 22 nn phng trnh (*)lun c hai nghim phn bit , do (d) v (P) lun ct nhau ti hai im phn bit Av B.b). A v B nm v hai pha ca trc tung phng trnh : x2 + mx + m 2 = 0 chai nghim tri du m 2 < 0 m < 2.
Bi 3 :
( )
( )
=++
=++
=++
327
)2(1111
19
xzyzxy
zyx
zyx
KX : .0,0,0 zyx
( ) ( )
( )
( ) ( )
2 2 2 2
2 2 2 2 2 2
2 2 2 2 2 2
2 2 2
2
2
2
81 2 81
81 2 27
2( ) 2 0
( ) ( ) ( ) 0
( ) 0( ) 0
( ) 0
x y z x y z xy yz zx
x y z xy yz zx x y z
x y z xy yz zx x y z xy yz zx
x y y z z x
x y x y
y z y z x y z
z xz x
+ + = + + + + + =
+ + = + + + + =
+ + = + + + + + + =
+ + =
= = = = = =
= =
Thay vo (1) => x = y = z = 3 .Ta thy x = y = z = 3 tha mn h phng trnh . Vy h phng trnh c nghim duynht x = y = z = 3.Bi 4:a). Xt ABM v NBM .Ta c: AB l ng knh ca ng trn (O)nn :AMB = NMB = 90o .M l im chnh gia ca cung nh ACnn ABM = MBN => BAM = BNM=> BAN cn nh B.T gic AMCB ni tip=> BAM = MCN ( cng b vi gc MCB).=> MCN = MNC ( cng bng gc BAM).=> Tam gic MCN cn nh Mb). Xt MCB v MNQ c :
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MC = MN (theo cm trn MNC cn ) ; MB = MQ ( theo gt) BMC = MNQ ( v : MCB = MNC ; MBC = MQN ).
=> )...( cgcMNQMCB = => BC = NQ .Xt tam gic vung ABQ c BQAC AB2 = BC . BQ = BC(BN + NQ)=> AB2 = BC .( AB + BC) = BC( BC + 2R)=> 4R2 = BC( BC + 2R) => BC = R)15(
Bi 5:
T :zyxzyx ++
=++1111
=> 01111
=++
++zyxzyx
=>( )
0=++
+++
+
zyxz
zzyx
xy
yx
( )( )
( )
( )( ) 0)(
0)(
011
2
=+++
=
++
++++
=
++++
xzzyyx
zyxxyz
xyzzyzxyx
zyxzxyyz
Ta c : x8 y8 = (x + y)(x-y)(x2+y2)(x4 + y4).=y9 + z9 = (y + z)(y8 y7z + y6z2 - .......... + z8)z10- x10 = (z + x)(z4 z3x + z2x2 zx3 + x4)(z5 - x5)
Vy M =4
3+ (x + y) (y + z) (z + x).A =
4
3
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7Bi 1: 1) Cho ng thng d xc nh bi y = 2x + 4. ng thng d/ i xng ving thng d qua ng thng y = x l:
A.y =2
1x + 2 ; B.y = x - 2 ; C.y =
2
1x - 2 ; D.y = - 2x - 4
Hy chn cu tr li ng.
2) Mt hnh tr c chiu cao gp i ng knh y ng y nc, nhng
chm vo bnh mt hnh cu khi ly ra mc nc trong bnh cn li3
2bnh. T s gia
bn knh hnh tr v bn knh hnh cu l A.2 ; B. 3 2 ; C. 3 3 ; D. mt kt qu khc.
Ba2: 1) Gii phng trnh: 2x4 - 11 x3 + 19x2 - 11 x + 2 = 02) Cho x + y = 1 (x > 0; y > 0) Tm gi tr ln nht ca A = x + y
Bi 3: 1) Tm cc s nguyn a, b, c sao cho a thc : (x + a)(x - 4) - 7Phn tch thnh tha s c : (x + b).(x + c)
2) Cho tam gic nhn xy, B, C ln lt l cc im c nh trn tia Ax, Ay saocho AB < AC, im M di ng trong gc xAy sao cho
MB
MA=
2
1
Xc nh v tr im M MB + 2 MC t gi tr nh nht.Bi 4: Cho ng trn tm O ng knh AB v CD vung gc vi nhau, ly im Ibt k trn oan CD.
a) Tm im M trn tia AD, im N trn tia AC sao cho I lag trung im caMN.
b) Chng minh tng MA + NA khng i.c) Chng minh rng ng trn ngoi tip tam gic AMN i qua hai im c
nh.
Hng dn
Bi 1: 1) Chn C. Tr li ng.2) Chn D. Kt qu khc: p s l: 1
Bi 2 : 1)A = (n + 1)4 + n4 + 1 = (n2 + 2n + 1)2 - n2 + (n4 + n2 + 1)= (n2 + 3n + 1)(n2 + n + 1) + (n2 + n + 1)(n2 - n + 1)= (n2 + n + 1)(2n2 + 2n + 2) = 2(n2 + n + 1)2
Vy A chia ht cho 1 s chnh phng khc 1 vi mi s nguyn dng n.
2) Do A > 0 nn A ln nht A2 ln nht.Xt A2 = ( x + y )2 = x + y + 2 xy = 1 + 2 xy (1)
Ta c:2
yx + xy (Bt ng thc C si)
=> 1 > 2 xy (2)
T (1) v (2) suy ra: A2 = 1 + 2 xy < 1 + 2 = 2
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M
D
C
B
A
x
K
O
N
M
I
D
C
BA
Max A2 = 2 x = y =2
1, max A = 2 x = y =
2
1
Bi3 Cu 1Vi mi x ta c (x + a)(x - 4) - 7 = (x + b)(x + c)Nn vi x = 4 th - 7 = (4 + b)(4 + c)C 2 trng hp: 4 + b = 1 v 4 + b = 7
4 + c = - 7 4 + c = - 1
Trng hp th nht cho b = - 3, c = - 11, a = - 10Ta c (x - 10)(x - 4) - 7 = (x - 3)(x - 11)Trng hp th hai cho b = 3, c = - 5, a = 2
Ta c (x + 2)(x - 4) - 7 = (x + 3)(x - 5)Cu2 (1,5im)
Gi D l im trn cnh AB sao cho:
AD =4
1AB. Ta c D l im c nh
MAB
MA=
2
1(gt) do
MA
AD=
2
1
Xt tam gic AMB v tam gic ADM c MB (chung)
AB
MA=
MA
AD=
21
Do AMB ~ ADM =>MD
MB=
AD
MA= 2
=> MD = 2MD (0,25 im)Xt ba im M, D, C : MD + MC > DC (khng i)Do MB + 2MC = 2(MD + MC) > 2DCDu "=" xy ra M thuc on thng DCGi tr nh nht ca MB + 2 MC l 2 DC
* Cch dng im M.
- Dng ng trn tm A bn knh21 AB
- Dng D trn tia Ax sao cho AD =4
1AB
M l giao im ca DC v ng trn (A;2
1AB)
Bi 4:a) Dng (I, IA) ct AD ti M ct tia AC ti NDo MN = 900 nn MN l ng knh
Vy I l trung im ca MNb) K MK // AC ta c : INC = IMK (g.c.g)
=> CN = MK = MD (v MKD vung cn)Vy AM+AN=AM+CN+CA=AM+MD+CA=> AM = AN = AD + AC khng i
c) Ta c IA = IB = IM = INVy ng trn ngoi tip AMN i qua hai im A, B c nh .
8Bi 1. Cho ba s x, y, z tho mn ng thi :
2 2 22 1 2 1 2 1 0 x y y z z x+ + = + + = + + =
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Tnh gi tr ca biu thc : 2007 2007 2007 A x y z= + + .
Bi 2). Cho biu thc : 2 25 4 2014 M x x y xy y= + + + .
Vi gi tr no ca x, y th M t gi tr nh nht ? Tm gi tr nh nht
Bi 3. Gii h phng trnh :
( ) ( )
2 2 18
1 . 1 72
x y x y
x x y y
+ + + = + + =
Bi 4. Cho ng trn tm O ng knh AB bn knh R. Tip tuyn ti im M bbt
k trn ng trn (O) ct cc tip tuyn ti A v B ln lt ti C v D.
a.Chng minh : AC . BD = R2.
b.Tm v tr ca im M chu vi tam gic COD l nh nht .
Bi 5.Cho a, b l cc s thc dng. Chng minh rng :
( )2
2 22
a ba b a b b a
++ + +
Bi 6).Cho tam gic ABC c phn gic AD. Chng minh : AD2 = AB . AC - BD . DC.
Hng dn giiBi 1. T gi thit ta c :
2
2
2
2 1 0
2 1 0
2 1 0
x y
y z
z x
+ + =
+ + =
+ + =
Cng tng v cc ng thc ta c : ( ) ( ) ( )2 2 22 1 2 1 2 1 0 x x y y z z+ + + + + + + + =
( ) ( ) ( )2 2 2
1 1 1 0 x y z + + + + + = 1 0
1 0
1 0
x
y
z
+ =
+ =
+ =
1 x y z = = =
( ) ( ) ( )2007 2007 20072007 2007 2007 1 1 1 3 A x y z = + + = + + = Vy : A = -3.
Bi 2.(1,5 im) Ta c :
( ) ( ) ( )2 24 4 2 1 2 2 2007 M x x y y xy x y= + + + + + + + +
( ) ( ) ( ) ( )2 2
2 1 2 1 2007 M x y x y= + + +
( ) ( ) ( )2
21 32 1 1 2007
2 4 M x y y
= + + +
Do ( )2
1 0y v ( ) ( )2
12 1 0
2x y
+
,x y
2007M min 2007 2; 1 M x y = = =
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Bi 3. t :( )
( )
1
1
u x x
v y y
= +
= +Ta c :
18
72
u v
uv
+ =
= u ; v l nghim ca phng
trnh :2
1 218 72 0 12; 6 X X X X + = = =
126uv
=
=; 612
uv
=
=
( )
( )
1 12
1 6
x x
y y
+ =
+ =;
( )
( )
1 6
1 12
x x
y y
+ =
+ =
Gii hai h trn ta c : Nghim ca h l :
(3 ; 2) ; (-4 ; 2) ; (3 ; -3) ; (-4 ; -3) v cc hon v.
Bi 4. a.Ta c CA = CM; DB = DM
Cc tia OC v OD l phn gic ca hai gc AOM v MOB nn OC OD
Tam gic COD vung nh O, OM l ng cao thuc cnh huyn CD nn :MO2 = CM . MD
R2 = AC . BD
b.Cc t gic ACMO ; BDMO ni tip
; MCO MAO MDO MBO = =
( ).COD AMB g g (0,25)
Do :1
. .
. .
Chu vi COD OM
Chu vi AMB MH= (MH1 AB)
Do MH1 OM nn1
1OM
MH
Chu vi COD chu vi AMB
Du = xy ra MH1 = OM M O M l im chnh gia ca cung AB
Bi 5 (1,5 im) Ta c :2 2
1 10; 0
2 2a b
a , b > 0
1 10; 0
4 4a a b b + +
1 1( ) ( ) 0
4 4a a b b + + + a , b > 0
1 02
a b a b + + + > Mt khc 2 0a b ab+ >
Nhn tng v ta c : ( ) ( ) ( )1
22
a b a b ab a b
+ + + +
( )( )2
2 22
a ba b a b b a
+ + + +
Bi 6. (1 im) V ng trn tm O ngoi tip ABC
oh
d
c
m
ba
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Gi E l giao im ca AD v (O)
Ta c: ABD CED (g.g)
. . BD AD
AB ED BD CD ED CD
= =
( )
2
. .
. .
AD AE AD BD CD
AD AD AE BDCD
=
=
Li c : ( ). ABD AEC g g
2
. .
. .
AB AD AB AC AE AD
AE AC
AD AB AC BDCD
= =
=
9
Cu 1: Cho hm s f(x) = 442 + xx
a) Tnh f(-1); f(5)
b) Tm x f(x) = 10
c) Rt gn A =4
)(2
x
xfkhi x 2
Cu 2: Gii h phng trnh
+=+
+=
)3)(72()72)(3(
)4)(2()2(
yxyx
yxyx
Cu 3: Cho biu thcA =
+
+
1:1
1
1
1
x
x
xx
x
x
xx
vi x > 0 v x
1
a) Rt gn A
b) Tm gi tr ca x A = 3
Cu 4: T im P nm ngoi ng trn tm O bn knh R, k hai tip tuyn PA; PB.
Gi H l chn ng vung gc h t A n ng knh BC.
a) Chng minh rng PC ct AH ti trung im E ca AH
b) Gi s PO = d. Tnh AH theo R v d.
Cu 5: Cho phng trnh 2x2 + (2m - 1)x + m - 1 = 0
Khng gii phng trnh, tm m phng trnh c hai nghim phn bit x 1; x2 tha
mn: 3x1 - 4x2 = 11
p n
Cu 1a) f(x) = 2)2(44 22 ==+ xxxx
c
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Suy ra f(-1) = 3; f(5) = 3
b)
=
=
=
==
8
12
102
10210)(
x
x
x
xxf
c))2)(2(
2
4
)(2
+
=
=
xx
x
x
xfA
Vi x > 2 suy ra x - 2 > 0 suy ra2
1
+=
xA
Vi x < 2 suy ra x - 2 < 0 suy ra2
1
+=
xA
Cu 2
( 2) ( 2)( 4) 2 2 4 8 4
( 3)(2 7) (2 7)( 3) 2 6 7 21 2 7 6 21 0
x y x y xy x xy y x x y
x y x y xy y x xy y x x y
= + = + = =
+ = + + = + + = =
x -2
y 2
Cu 3 a) Ta c: A =
+
+
1:1
1
1
1
x
xx
x
x
x
xx
=
+
+
++
11
)1(:
1
1
)1)(1(
)1)(1(
x
x
x
xx
x
x
xx
xxx=
+
+
1:
1
1
1
1
x
xxx
x
x
x
xx=
1:
1
11
++
x
x
x
xxx=
1:
1
2
+
x
x
x
x=
x
x
x
x 1
1
2
+=
x
x2
b) A = 3 =>x
x2= 3 => 3x + x - 2 = 0 => x = 2/3
Cu 4
Do HA // PB (Cng vung gc vi BC)
a) nn theo nh l Ta let p dng cho CPB ta c
CB
CH
PB
EH= ; (1)
Mt khc, do PO // AC (cng vung gc vi AB)
=> POB = ACB (hai gc ng v)
=> AHC POB
Do :OB
CH
PB
AH= (2)
OB CH
E
A
P
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Do CB = 2OB, kt hp (1) v (2) ta suy ra AH = 2EH hay E l trung im ca
AH.
b) Xt tam gic vung BAC, ng cao AH ta c AH2 = BH.CH = (2R - CH).CH
Theo (1) v do AH = 2EH ta c
.)2( 2PBAH.CB2PBAH.CBAH2 = R
AH2.4PB2 = (4R.PB - AH.CB).AH.CB
4AH.PB2 = 4R.PB.CB - AH.CB2
AH (4PB2 +CB2) = 4R.PB.CB
2
222
222
222
2222
d
Rd.2.R
4R)R4(d
Rd.8R
(2R)4PB
4R.2R.PB
CB4.PB
4R.CB.PBAH
=
+
=
+=
+=
Cu 5 phng trnh c 2 nghim phn bit x1 ; x2 th > 0
(2m - 1)2 - 4. 2. (m - 1) > 0
T suy ra m 1,5 (1)
Mt khc, theo nh l Vit v gi thit ta c:
=
=
=+
114x3x2
1m
.xx
2
12mxx
21
21
21
=
=
=
118m-26
77m4
7
4m-133
8m-26
77m
x
7
4m-13x
1
1
Gii phng trnh 118m-26
77m4
7
4m-133 =
ta c m = - 2 v m = 4,125 (2)
i chiu iu kin (1) v (2) ta c: Vi m = - 2 hoc m = 4,125 th phng
trnh cho c hai nghim phn bit tha mn: x1 + x2 = 11
10
Cu 1: Cho P = 21
x
x x
+
+
1
1
x
x x
+
+ +-
11
x
x
+
a/. Rt gn P.
b/. Chng minh: P BC. im D di ng trn cnh AB, ( D khngtrng vi A, B). Gi (O) l ng trn ngoi tip BCD . Tip tuyn ca (O) ti C v Dct nhau K .
a/. Chng minh t gic ADCK ni tip.b/. T gic ABCK l hnh g? V sao?
c/. Xc nh v tr im D sao cho t gic ABCK l hnh bnh hnh.
p nCu 1:iu kin: x 0 v x 1. (0,25 im)
P =2
1
x
x x
+
+
1
1
x
x x
+
+ +-
1
( 1)( 1)
x
x x
+
+
=3
2
( ) 1
x
x
+
+
1
1
x
x x
+
+ +-
1
1x
=2 ( 1)( 1) ( 1)
( 1)( 1)
x x x x x
x x x
+ + + + +
+ +
=( 1)( 1)
x x
x x x
+ +=
1
x
x x+ +
b/. Vi x 0 v x 1 .Ta c: P 0
( x - 1)2
> 0. ( ng v x 0 v x 1)
Cu 2:a/. Phng trnh (1) c nghim khi v ch khi 0. (m - 1)2 m2 3 0 4 2m 0 m 2.b/. Vi m 2 th (1) c 2 nghim.Gi mt nghim ca (1) l a th nghim kia l 3a . Theo Viet ,ta c:
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23 2 2
.3 3
a a m
a a m
+ =
=
a=1
2
m 3(
12
m )2 = m2 3
m2 + 6m 15 = 0 m = 3 2 6 ( tha mn iu kin).
Cu 3:iu kin x 0 ; 2 x2 > 0 x 0 ; x < 2 .
t y = 22 x > 0
Ta c:
2 2 2 (1)
1 12 (2)
x y
x y
+ =
+ =
T (2) c : x + y = 2xy. Thay vo (1) c : xy = 1 hoc xy = -1
2
* Nu xy = 1 th x+ y = 2. Khi x, y l nghim ca phng trnh:
X2 2X + 1 = 0 X = 1 x = y = 1.* Nu xy = -
1
2th x+ y = -1. Khi x, y l nghim ca phng trnh:
X2 + X -1
2= 0 X =
1 3
2
V y > 0 nn: y =1 3
2 +
x =1 3
2
Vy phng trnh c hai nghim: x1 = 1 ; x2 =1 3
2
Cu 4: c/. Theo cu b, t gic ABCK l hnh thang.
Do , t gic ABCK l hnh bnh hnh AB // CK BAC ACK=
M1
2ACK= sEC =
1
2sBD = DCB
Nn BCD BAC= Dng tia Cy sao cho BCy BAC= .Khi , D l giao im ca AB v Cy.
Vi gi thit AB > BC th BCA > BAC > BDC. D AB .
Vy im D xc nh nh trn l im cn tm.
11Cu 1: a) Xc nh x R biu thc :A =
xxxx
+
+
1
11
2
2 L mt s t nhin
O
D
CB
A
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b. Cho biu thc: P =22
2
12 +++
+++
++ zzx
z
yyz
y
xxy
xBit x.y.z = 4 , tnh
P .Cu 2:Cho cc im A(-2;0) ; B(0;4) ; C(1;1) ; D(-3;2)
a. Chng minh 3 im A, B ,D thng hng; 3 im A, B, C khng thng hng.b. Tnh din tch tam gic ABC.
Cu3 Gii phng trnh: 521 3 = xx Cu 4 Cho ng trn (O;R) v mt im A sao cho OA = R 2 . V cc tip tuynAB, AC vi ng trn. Mt gc xOy = 450 ct on thng AB v AC ln lt ti Dv E.
Chng minh rng:a.DE l tip tuyn ca ng trn ( O ).
b. RDER 0 v
2=xyz
Nhn c t v mu ca hng t th 2 vi x ; thay 2 mu ca hng t th 3 bi xyz ta c:
P = 12
2
2(
2
22=
++
++=
+++
+++
++ xxy
xyx
xyxz
z
xxy
xy
xxy
x(1)
1=P v P > 0Cu 2: a.ng thng i qua 2 im A v B c dng y = ax + bim A(-2;0) v B(0;4) thuc ng thng AB nn b = 4; a = 2Vy ng thng AB l y = 2x + 4.im C(1;1) c to khng tho mn y = 2x + 4 nn C khng thuc ng thng AB
A, B, C khng thng hng.im D(-3;2) c to tho mn y = 2x + 4 nn im D thuc ng thng AB A,B,D thng hn
b.Ta c :AB2 = (-2 0)2 + (0 4)2 =20AC2 = (-2 1)2 + (0 1)2 =10BC2 = (0 1)2 + (4 1)2 = 10
AB2 = AC2 + BC2 ABC vung ti C
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Vy SABC = 1/2AC.BC = 510.1021
= ( n v din tch )
Cu 3: kx x 1, t vxux == 3 2;1 ta c h phng trnh:
=+
=
1
532 vu
vu
Gii h phng trnh bng phng php th ta c: v = 2 x = 10.
Cu 4a.p dng nh l Pitago tnh cAB = AC = R ABOC l hnhvung (0.5)K bn knh OM sao choBOD = MOD MOE = EOC (0.5)Chng minh BOD = MOD
OMD = OBD = 900Tng t: OME = 900 D, M, E thng hng. Do DE l tip tuyn ca ng trn (O).
b.Xt ADE c DE < AD +AE m DE = DB + EC 2ED < AD +AE +DB + EC hay 2DE < AB + AC = 2R DE < RTa c DE > AD; DE > AE ; DE = DB + EC
Cng tng v ta c: 3DE > 2R DE >3
2R
Vy R > DE >3
2R
12Cu 1: Cho hm s f(x) = 442 + xx
a) Tnh f(-1); f(5)
b) Tm x f(x) = 10
c) Rt gn A =4
)(2
x
xfkhi x 2
Cu 2: Gii h phng trnh
+=+
+=
)3)(72()72)(3(
)4)(2()2(
yxyx
yxyx
Cu 3: Cho biu thc A =
+
+
1:
1
11
1
x
xx
x
x
x
xxvi x > 0 v x 1
a) Rt gn A
B
MA
O
C
D
E
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2) Tm gi tr ca x A = 3
Cu 4: T im P nm ngoi ng trn tm O bn knh R, k hai tip tuyn PA; PB.
Gi H l chn ng vung gc h t A n ng knh BC.
a) Chng minh rng PC ct AH ti trung im E ca AH
b) Gi s PO = d. Tnh AH theo R v d.Cu 5: Cho phng trnh 2x2 + (2m - 1)x + m - 1 = 0
Khng gii phng trnh, tm m phng trnh c hai nghim phn bit x 1; x2 tha
mn: 3x1 - 4x2 = 11
p n
Cu 1
a) f(x) = 2)2(44 22 ==+ xxxx
Suy ra f(-1) = 3; f(5) = 3
b)
=
=
=
==
8
12
102
10210)(
x
x
x
xxf
c))2)(2(
2
4
)(2
+
=
=
xx
x
x
xfA
Vi x > 2 suy ra x - 2 > 0 suy ra2
1
+=
xA
Vi x < 2 suy ra x - 2 < 0 suy ra 21+= xA
Cu 2
=
=
=+
=
+=+
+=
+=+
+=
2y
-2x
0
4
2167221762
8422
)3)(72()72)(3(
)4)(2()2(
yx
yx
xyxyxyxy
xyxyxxy
yxyx
yxyx
Cu 3a) Ta c: A =
+
+
1:
1
1
1
1
x
xx
x
x
x
xx
=
+
+
++
11
)1(:
1
1
)1)(1(
)1)(1(
x
x
x
xx
x
x
xx
xxx
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=
+
+
1:
1
1
1
1
x
xxx
x
x
x
xx
=1
:1
11
++
x
x
x
xxx
=1
:12
+
xx
xx =
xx
xx 1
12
+ =x
x2
b) A = 3 =>x
x2= 3 => 3x + x - 2 = 0 => x = 2/3
Cu 4
a) Do HA // PB (Cng vung gc vi BC)
b) nn theo nh l Ta let p dng cho tam gic CPB ta c
CB
CH
PB
EH= ; (1)
Mt khc, do PO // AC (cng vung gc vi AB)
=> POB = ACB (hai gc ng v)
=> AHC POB
Do :OB
CH
PB
AH= (2)
Do CB = 2OB, kt hp (1) v (2) ta suy ra AH = 2EH hay E l trug im ca
AH.
b) Xt tam gic vung BAC, ng cao AH ta c AH2 = BH.CH = (2R - CH).CH
Theo (1) v do AH = 2EH ta c
.)2(2PB
AH.CB
2PB
AH.CBAH2 = R
O
B CH
E
A
P
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AH2.4PB2 = (4R.PB - AH.CB).AH.CB
4AH.PB2 = 4R.PB.CB - AH.CB2
AH (4PB2 +CB2) = 4R.PB.CB
2
222
222
222
2222
d
Rd.2.R
4R)R4(d
Rd.8R
(2R)4PB
4R.2R.PB
CB4.PB
4R.CB.PBAH
=
+
=
+=
+=
Cu 5 (1)
phng trnh c 2 nghim phn bit x1 ; x2 th > 0
(2m - 1)2 - 4. 2. (m - 1) > 0
T suy ra m 1,5 (1)
Mt khc, theo nh l Vit v gi thit ta c:
=
=
=+
114x3x2
1m.xx
2
12mxx
21
21
21
=
=
=
118m-26
77m4
7
4m-133
8m-26
77mx
7
4m-13x
1
1
Gii phng trnh 11
8m-26
77m4
7
4m-133 =
ta c m = - 2 v m = 4,125 (2)
i chiu iu kin (1) v (2) ta c: Vi m = - 2 hoc m = 4,125 th phng trnh cho c hai nghim phn bit t
13Cu I :Tnh gi tr ca biu thc:
A = 531
+ + 751
+ + 971
+ + .....+ 99971
+
B = 35 + 335 + 3335 + ..... + 43421399
35.....3333s
Cu II :Phn tch thnh nhn t :1)X2 -7X -182) (x+1) (x+2)(x+3)(x+4)3) 1+ a5 + a10
Cu III :
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1)Chng minh : (ab+cd)2 (a2+c2)( b2 +d2)2)p dng : cho x+ 4y = 5 . Tm GTNN ca biu thc : M= 4x2 + 4y2
Cu 4 : Cho tam gic ABC ni tip ng trn (O), I l trung im ca BC, M l mtim trn on CI ( M khc C v I ). ng thng AM ct (O) ti D, tip tuyn cang trn ngoi tip tam gic AIM ti M ct BD v DC ti P v Q.
a) Chng minh DM.AI= MP.IB
b) Tnh t s :MQMP
Cu 5:
Cho P =x
xx
+
1
342
Tm iu kin biu thc c ngha, rt gn biu thc.
p n
Cu 1 :
1) A = 53
1
+ + 75
1
+ + 97
1
+ + .....+ 9997
1
+
=2
1( 35 + 57 + 79 + .....+ 9799 ) =
2
1( 399 )
2) B = 35 + 335 + 3335 + ..... + 43421399
35.....3333s
=
=33 +2 +333+2 +3333+2+.......+ 333....33+2= 2.99 + ( 33+333+3333+...+333...33)
= 198 +3
1( 99+999+9999+.....+999...99)
198 +
3
1( 102 -1 +103 - 1+104 - 1+ ....+10100 1) = 198 33 +
B =
27
1010 2101+165
Cu 2: 1)x2 -7x -18 = x2 -4 7x-14 = (x-2)(x+2) - 7(x+2) = (x+2)(x-9) (1)2)(x+1)(x+2)(x+3)(x+4) -3= (x+1)(x+4)(x+2)(x+3)-3
= (x2+5x +4)(x2 + 5x+6)-3= [x2+5x +4][(x2 + 5x+4)+2]-3= (x2+5x +4)2 + 2(x2+5x +4)-3=(x2+5x +4)2 - 1+ 2(x2+5x +4)-2= [(x2+5x +4)-1][(x2+5x +4)+1] +2[(x2+5x +4)-1]= (x2+5x +3)(x2+5x +7)
3) a10+a5+1= a10+a9+a8+a7+a6 + a5 +a5+a4+a3+a2+a +1- (a9+a8+a7 )- (a6 + a5 +a4)- ( a3+a2+a )= a8(a2 +a+1) +a5(a2 +a+1)+ a3(a2 +a+1)+ (a2 +a+1)-a7(a2 +a+1)
-a4(a2 +a+1)-a(a2 +a+1)=(a2 +a+1)( a8-a7+ a5 -a4+a3 - a +1)
Cu 3: 41) Ta c : (ab+cd)2 (a2+c2)( b2 +d2)
a2b2+2abcd+c2d2 a2b2+ a2d2 +c2b2 +c2d2
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0 a2d2 - 2cbcd+c2b2 0 (ad - bc)2 (pcm )
Du = xy ra khi ad=bc.2) p dng hng ng thc trn ta c :52 = (x+4y)2 = (x. + 4y) (x2 + y2) )161( + =>
x2 + y2 17
25=> 4x2 + 4y2
17
100du = xy ra khi x=
17
5, y =
17
20(2)
Cu 4 : 5Ta c : gc DMP= gc AMQ = gc AIC. Mt khc gc ADB = gc BCA=>
MPD ng dng vi ICA =>IA
MP
CI
DM= => DM.IA=MP.CI hay DM.IA=MP.IB
(1).Ta c gc ADC = gc CBA,
Gc DMQ = 1800 - AMQ=1800 - gc AIM = gc BIA.Do DMQ ng dng vi BIA =>
IA
MQ
BI
DM= => DM.IA=MQ.IB (2)
T (1) v (2) ta suy raMQ
MP= 1
Cu 5 P xc nh th : x2-4x+3 0 v 1-x >0
T 1-x > 0 => x < 1Mt khc : x2-4x+3 = (x-1)(x-3), V x < 1 nn ta c :(x-1) < 0 v (x-3) < 0 t suy ra tch ca (x-1)(x-3) > 0Vy vi x < 1 th biu thc c ngha.
Vi x < 1 Ta c :
P = xxx
+
1
342
= xx
xx
=
31
)3)(1(
14
Cu 1 : a. Rt gn biu thc .( )22 1
111
+++=
aaA Vi a > 0.
b. Tnh gi tr ca tng.222222 100
1
99
11...
3
1
2
11
2
1
1
11 +++++++++=B
Cu 2 : Cho pt 012 =+ mmxx
a. Chng minh rng pt lun lun c nghim vi m .
b. Gi 21, xx l hai nghim ca pt. Tm GTLN, GTNN ca bt.
( )12
32
212
22
1
21
+++
+=
xxxx
xxP
Cu 3 : Cho 1,1 yx Chng minh.
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xyyx +
++
+ 12
1
1
1
122
Cu 4 Cho ng trn tm O v dy AB. M l im chuyn ng trn ng trn,t M k MH AB (H AB). Gi E v F ln lt l hnh chiu vung gc ca H trnMA v MB. Qua M k ng thng vung gc vi EF ct dy AB ti D.
1. Chng minh rng ng thng MD lun i qua 1 im c nh khi M thay itrn ng trn.
2. Chng minh.
BH
AD
BD
AH
MB
MA.
2
2
=
Hng dn
Cu 1 a. Bnh phng 2 v ( )112
+
++=
aa
aa
A (V a > 0).c. p dng cu a.
100
9999
100
1100
1
111
==
++=
B
aaA
Cu 2 a. : cm m 0 B (2 ) p dng h thc Viet ta c:
=
=+
121
21
mxx
mxx
212
2+
+
= m
mP (1) Tm k pt (1) c nghim theo n.
11
221
121
==
==
mGTNN
mGTLN
P
Cu 3 : Chuyn v quy ng ta c.
bt( )
( )( )
( )
( )( )
0
1111
22
++
+
++
xyy
yxy
xyx
xyx
( ) ( ) 012 xyyx ng v 1xy
Cu 4: a- K thm ng ph.- Chng minh MD l ng knh ca (o)=> ........b.
Gi E', F' ln lt l hnh chiu ca D trn MA v MB.
oE'
E
A
F
F'
B
I
D
H
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t HE = H1HF = H2
( )1..
...
22
21
MBhHF
MAhHE
BH
AD
BD
AH=
HEF ''EDF hHEhHF .. 2 =
Thay vo (1) ta c:BH
AD
BD
AH
MB
MA.
2
2
=
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15
Cu 1: Cho biu thc D =
+
++
+
ab
ba
ab
ba
11:
+++
ab
abba
1
21
a) Tm iu kin xc nh ca D v rt gn Db) Tnh gi tr ca D vi a =
32
2
c) Tm gi tr ln nht ca D
Cu 2: Cho phng trnh32
2
x2- mx +
32
2
m2 + 4m - 1 = 0 (1)
a) Gii phng trnh (1) vi m = -1
b) Tm m phng trnh (1) c 2 nghim tho mn 2121
11xx
xx+=+
Cu 3: Cho tam gic ABC ng phn gic AI, bit AB = c, AC = b,
)90( 0== A Chng minh rng AI =cb
Cosbc
+
2.2
(Cho Sin2 CosSin2= )
Cu 4: Cho ng trn (O) ng knh AB v mt im N di ng trn mt na ng
trn sao cho .BNAN))
V vo trong ng trn hnh vung ANMP.
a) Chng minh rng ng thng NP lun i qua im c nh Q.
b) Gi I l tm ng trn ni tip tam gic NAB. Chng minh t gic ABMI ni
tip.
c) Chng minh ng thng MP lun i qua mt im c nh.
Cu 5: Cho x,y,z; xy + yz + zx = 0 v x + y + z = -1
Hy tnh gi tr ca:
B =x
xyz
y
zx
z
xy++
p n
Cu 1: a) - iu kin xc nh ca D l
1
0
0
ab
b
a
- Rt gn D
D =
+
ab
aba
1
22:
++
ab
abba
1
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c
ba
I
CB
A
2
2
D =1
2
+a
a
b) a = 13)13(1
32(2
32
2 2+=+=
+=
+a
Vy D = 34232
132
2322
=+
+
c) p dng bt ng thc cauchy ta c112 + Daa
Vy gi tr ca D l 1
Cu 2: a) m = -1 phng trnh (1) 09202
9
2
1 22=+=+ xxxx
+=
=
101
101
2
1
x
x
b) phng trnh 1 c 2 nghim th410280 + mm (*)
+ phng trnh c nghim khc 0
+
+
234
234
01421
2
1
2
m
m
mm
(*)
+
=
=+=++=+
01
00)1)((
11
21
21212121
21 xx
xxxxxxxx
xx
+=
=
=
=+
=
194194
0
038
022
m
m
m
mm
m
Kt hp vi iu kin (*)v (**) ta c m = 0 v 194 =m Cu 3:
+ ;2
.2
1 cSinAIS
ABI=
+ ;2
.2
1 bSinAIS AIC =
+ ;21
bcSinS ABC = AICABIABC SSS +=
cb
bcCos
cbSin
bcSinAI
cbAISinbcSin
+=
+
=
+=
22
)(2
)(2
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1
2
1
2
1
F
I
Q
P
N
M
BA
Cu 4: a) 21 NN = Gi Q = NP )(O
QA QB =) )
Suy ra Q c nh
b) )( 211 AMA == T gic ABMI ni tipc) Trn tia i ca QB ly im F sao cho QF = QB, F c nh.Tam gic ABF c: AQ = QB = QF ABF vung ti A 00 4545 == BFAB
Li c == 10
145 PAFBP T gic APQF ni tip
090 == FQAFPA
Ta c: 000 1809090 =+=+ MPAFPA M1,P,F Thng hng
Cu 5: Bin i B = xyz
++
222
111
zyx= 2
2. ==xyz
xyzL
16
Bi 1: Cho biu thc A =2
4( 1) 4( 1) 1. 1
14( 1)
x x x x
xx x
+ +
a) Tm iu kin ca x A xc nhb) Rt gn A
Bi 2 : Trn cng mt mt phng ta cho hai im A(5; 2) v B(3; -4)a) Vit phng tnh ng thng ABb) Xc nh im M trn trc honh tam gic MAB cn ti M
Bi 3 : Tm tt c cc s t nhin m phng trnh n x sau:x2 - m2x + m + 1 = 0
c nghim nguyn.Bi 4 : Cho tam gic ABC. Phn gic AD (D BC) v ng trn tm O qua A v Dng thi tip xc vi BC ti D. ng trn ny ct AB v AC ln lt ti E v F.Chng minh
a) EF // BCb) Cc tam gic AED v ADC; D v ABD l cc tam gic ng dng.
c) AE.AC = .AB = AC2Bi 5 : Cho cc s dng x, y tha mn iu kin x2 + y2 x3 + y4. Chng minh:
x3 + y3 x2 + y2 x + y 2
p nBi 1:a) iu kin x tha mn
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2
1 0
4( 1) 0
4( 1) 0
4( 1) 0
x
x x
x x
x x
+
>
1
1
1
2
x
x
x
x
x > 1 v x 2
KL: A xc nh khi 1 < x < 2 hoc x > 2
b) Rt gn A
A =2 2
2
( 1 1) ( 1 1) 2.
1( 2)
x x x
xx
+ +
A =1 1 1 1 2
.2 1
x x x
x x
+ +
Vi 1 < x < 2 A =2
1 x
Vi x > 2 A =2
1x
Kt lunVi 1 < x < 2 th A =
2
1 x
Vi x > 2 th A =2
1x
Bi 2:a) A v B c honh v tung u khc nhau nn phng trnh ng thng AB cdng y = ax + b
A(5; 2) AB 5a + b = 2B(3; -4) AB 3a + b = -4
Gii h ta c a = 3; b = -13Vy phng trnh ng thng AB l y = 3x - 13
b) Gi s M (x, 0) xx ta c
MA = 2 2( 5) (0 2)x +
MB = 2 2( 3) (0 4)x + +
MAB cn MA = MB 2 2( 5) 4 ( 3) 16x x + = +
(x - 5)2 + 4 = (x - 3)2 + 16 x = 1Kt lun: im cn tm: M(1; 0)
Bi 3:Phng trnh c nghim nguyn khi = m4 - 4m - 4 l s chnh phngTa li c: m = 0; 1 th < 0 loim = 2 th = 4 = 22 nhnm 3 th 2m(m - 2) > 5 2m2 - 4m - 5 > 0 - (2m2 - 2m - 5) < < + 4m + 4 m4 - 2m + 1 < < m4 (m2 - 1)2 < < (m2)2
FE
A
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khng chnh phngVy m = 2 l gi tr cn tm.
Bi 4:
a) 1( )2
EAD EFD sdED= = (0,25)
1( )
2
FAD FDC sdFD= = (0,25)
m EDA FAD EFD FDC= = (0,25) EF // BC (2 gc so le trong bng nhau)
b) AD l phn gic gc BAC nn DE DF= s
1
2ACD = s(AED DF ) =
1
2sAE = sADE
do ACD ADE= v EAD DAC= DADC (g.g)
Tng t: s1 1
( )2 2
ADF sdAF sd AFD DF= = =1
( )2
sdAFD DE sdABD = ADF ABD=
do AFD ~ (g.gc) Theo trn:
+ AED ~ DB
AE AD
AD AC= hay AD2 = AE.AC (1)
+ ADF ~ ABD AD AF
AB AD=
AD2 = AB.AF (2)T (1) v (2) ta c AD2 = AE.AC = AB.AF
Bi 5 (1):
Ta c (y2
- y) + 2 0 2y3
y4
+ y2
(x3 + y2) + (x2 + y3) (x2 + y2) + (y4 + x3)
m x3 + y4 x2 + y3 do x3 + y3 x2 + y2 (1)
+ Ta c: x(x - 1)2 0: y(y + 1)(y - 1)2 0 x(x - 1)2 + y(y + 1)(y - 1)2 0 x3 - 2x2 + x + y4 - y3 - y2 + y 0 (x2 + y2) + (x2 + y3) (x + y) + (x3 + y4)
m x2 + y3 x3 + y4 x2 + y2 x + y (2)
v (x + 1)(x - 1) 0. (y - 1)(y3 -1) 0x3 - x2 - x + 1 + y4 - y - y3 + 1 0
(x + y) + (x2 + y3) 2 + (x3 + y4)m x2 + y3 x3 + y4
x + y 2T (1) (2) v (3) ta c:
x3 + y3 x2 + y2 x + y 2
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14
Cu 1: x- 4(x-1) + x + 4(x-1) 1cho A= ( 1 - )
x2- 4(x-1) x-1
a/ Rt gn biu thc A.b/ Tm gi tr nguyn ca x A c gi tr nguyn.Cu 2: Xc nh cc gi tr ca tham s m phng trnh
x2- (m+5)x- m + 6 = 0C 2 nghim x1 v x2 tho mn mt trong 2 iu kin sau:a/ Nghim ny ln hn nghim kia mt n v.b/ 2x1+3x2=13Cu 3 Tm gi tr ca m h phng trnh
mx-y=1m3x+(m2-1)y =2
v nghim, v s nghim.Cu 4: Tm max v min ca biu thc: x2+3x+1x2+1
Cu 5: T mt nh A ca hnh vung ABCD k hai tia to vi nhau mt gc 450. Mttia ct cnh BC ti E ct ng cho BD ti P. Tia kia ct cnh CD ti F v ct ngcho BD ti Q.a/ Chng minh rng 5 im E, P, Q, F v C cng nm trn mt ng trn.b/ Chng minh rng: SAEF=2SAQPc/ K trung trc ca cnh CD ct AE ti M tnh s o gc MAB bit CPD=CM
Hng dnCu 1: a/ Biu thc A xc nh khi x2 v x>1
( x-1 -1)2+ ( x-1 +1)2 x-2A= . ( )
(x-2)2 x-1x- 1 -1 + x-1 + 1 x- 2 2 x- 1 2
= . = =x-2 x-1 x-1 x-1
b/ A nguyn th x- 1 l c dng ca 1 v 2* x- 1 =1 th x=0 loi* x- 1 =2 th x=5vy vi x = 5 th A nhn gi tr nguyn bng 1Cu 2: Ta c x = (m+5)2-4(-m+6) = m2+14m+10 phng trnhc hai nghimphnbit khi vch khi m-7-4 3 v m-7+4 3 (*)a/ Gi s x2>x1 ta c h x2-x1=1
(1)x1+x2=m+5
(2)
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11
Q
PM
F
E
D C
BA
x1x2 =-m+6(3)
Gii h tac m=0 v m=-14 tho mn (*)b/ Theo gi thit ta c: 2x1+3x2 =13
(1)x1+x2 = m+5
(2)x1x2 =-m+6
(3)gii h ta c m=0 v m= 1 Tho mn (*)
Cu 3:* h v nghim th m/m3
=-1/(m2-1) 1/23m3-m=-m3 m2(4m2- 1)=0 m=0 m=03m2-1-2 3m2-1 m=1/2 m=1/2
m*Hv s nghim th: m/m3=-1/(m2-1) =1/2
3m3-m=-m3 m=03m2-1= -2 m=1/2
V nghimKhng c gi tr no ca m h v s nghim.
Cu 4: Hm s xc nh vi x(v x2+10) x2+3x+1gi y
0
l 1 gi trca hmphng trnh: y0
=x2+1
(y0-1)x2-6x+y0-1 =0 c nghim
*y0=1 suy ra x = 0 y0 1; =9-(y0-1)20 (y0-1)
29 suy ra-2 y0 4Vy: ymin=-2 v y max=4Cu 5: ( Hc sinh t v hnh)Giia/ A1 v B1 cng nhn on QE di mt gc 45
0 t gic ABEQ ni tip c.
FQE = ABE =1v.chng minh tng t ta c FBE = 1v Q, P, C cng nm trn ng trn ng kinh EF.b/ T cu a suy ra AQE vung cn.
AE
AQ= 2 (1)
tng t APF cng vung cn
AF
AB= 2 (2)
t (1) v (2) AQP ~ AEF (c.g.c)AEF
AQP
S
S= ( 2 )2 hay SAEF = 2SAQP
c/ thy CPMD ni tip, MC=MD v APD= CPD MCD= MPD= APD= CPD= CMDMD=CD MCD u MPD=600m MPD l gc ngoi ca ABM ta c APB=450 vy MAB=600- 450 =150
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17
Bi 1: Cho biu thc M = xx
x
x
xx
x
+
+
+
++
2
3
3
12
65
92
a. Tm iu kin ca x M c ngha v rt gn Mb. Tm x M = 5c. Tm x Z M Z.bi 2: a) Tm x, y nguyn dng tho mn phng trnh
3x2 +10 xy + 8y2 =96
b) Tm x, y bit / x - 2005/ + /x - 2006/ +/y - 2007/+/x- 2008/ = 3
Bi 3: a. Cho cc s x, y, z dng tho mnx
1 +y
1+
z
1 = 4
Chng ming rng: zyx ++2
1
+ zyx ++ 2
1
+ zyx 2
1
++ 1
b. Tm gi tr nh nht ca biu thc: B =2
2 20062
x
xx +(vi x 0 )
Bi 4: Cho hnh vung ABCD. K tia Ax, Ay sao cho yAx = 45 0 Tia Ax ct CB v BD ln lt ti E v P, tia Ay ct CD v BD ln lt ti F v Qa. Chng minh 5 im E; P; Q; F; C cng nm trn mt ng trnb. S AEF = 2 S APQ
K ng trung trc ca CD ct AE ti M. Tnh s o gc MAB bit DPC = DMC Bi 5: (1)
Cho ba s a, b , c khc 0 tho mn: 0111 =++ cba ; Hy tnh P = 222 bac
a
bc
c
ac++
p n
Bi 1:M =x
x
x
x
xx
x
++
++
+
2
3
3
12
65
92
a.K 9;4;0 xxx 0,5
Rt gn M =( )( )32
2123392
+++
xx
xxxxx
Bin i ta c kt qu: M =( )( )32
2
xx
xxM =
( )( )( )( ) 3
1
23
21
+=
+
x
xM
xx
xx
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( )
1644
16
416
1551
351
53
15M.b.
===
=
=+
=+
=
=
xx
x
xx
xx
x
x
c. M =3
41
3
43
3
1
+=
+=
+
xx
x
x
x
Do M z nn 3x l c ca 4 3x nhn cc gi tr: -4; -2; -1; 1; 2; 4
{ }49;25;16;4;1 x do 4x { }49;25;16;1x
Bi 2 a. 3x2
+ 10xy + 8y2
= 96 3x2 + 4xy + 6xy + 8y2 = 96
(3x2 + 6xy) + (4xy + 8y2) = 96
3x(x + 2y) + 4y(x +2y) = 96
(x + 2y)(3x + 4y) = 96
Do x, y nguyn dng nn x + 2y; 3x + 4y nguyen dng v 3x + 4y > x + 2y 3
m 96 = 25. 3 c cc c l: 1; 2; 3; 4; 6; 8; 12; 24; 32; 48; 96 c biu din thnh
tch 2 tha s khng nh hn 3 l: 96 = 3.32 = 4.24 = 6. 16 = 8. 12Li c x + 2y v 3x + 4y c tch l 96 (L s chn) c tng 4x + 6y l s chn
do
=+
=+
2443
62
yx
yxH PT ny v nghim
Hoc
=+
=+
1643
62
yx
yx
=
=
1
4
y
x
Hoc
=+
=+
1243
82
yx
yxH PT v nghim
Vy cc s x, y nguyn dng cn tm l (x, y) = (4, 1)
b. ta c /A/ = /-A/ AA
Nn /x - 2005/ + / x - 2006/ = / x - 2005/ + / 2008 - x/
3/3//20082005/ =+ xx (1)
m /x - 2005/ + / x - 2006/ + / y - 2007/ + / x - 2008/ = 3 (2)
Kt hp (1 v (2) ta c / x - 2006/ + / y - 2007/ 0 (3)
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(3) sy ra khi v ch khi
=
=
=
=
2007
2006
0/2007/
0/2006/
y
x
y
x
Bi 3a. Trc ht ta chng minh bt ng thc ph
b. Vi mi a, b thuc R: x, y > 0 ta c
( )
(*)
222
yx
ba
y
b
x
a
+
+
+
(a2y + b2x)(x + y) ( ) xyba 2+
a2y2 + a2xy + b2 x2 + b2xy a2xy + 2abxy + b2xy
a2y2 + b2x2 2abxy
a2y2 2abxy + b2x2 0
(ay - bx)2 0 (**) bt ng thc (**) ng vi mi a, b, v x,y > 0
Du (=) xy ra khi ay = bx hay
a b
x y=
p dung bt ng thc (*) hai ln ta c2 2 2 2 2
1 1 1 1 1 1 1 11 2 2 2 2 4 4 4 4
2 2x y z x y z x y x z x y x z
+ + +
= + = +
+ + + + + + + +
2 2 2 21 1 1 1
1 2 1 14 4 4 416 x y x z x y z
+ + + = + +
Tng t1 1 1 2 1
2 16x y z x y z
+ +
+ +
1 1 1 1 2
2 16x y z x y z
+ +
+ +
Cng tng v cc bt ng thc trn ta c:
1 1 1 1 2 1 1 1 1 2 1 1 1 1 2
2 2 2 16 16 16
1 4 4 4 4 1 1 1 1 .4 116 16 4
x y z x y z x y z x y z x y z x y z
x y z x y z
+ + + + + + + + + +
+ + + + + +
+ + + + =
V1 1 1
4 x y z
+ + =
( )2
2
2 20060
x xB x
x
+=
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Ta c:x
xxB
x
xxB
2006
20062006.2200620062 22
2
2+
=+
=
( ) ( )
2006
2005
2006
20052006200520062
2
2
22
++
+
=x
x
x
xxB
V (x - 2006)2 0 vi mi x
x2 > 0 vi mi x khc 0
( )2
2
2006 2005 20050 2006
2006 2006 2006
x B B khix
x
= =
Bi 4a. 045 EBQ EAQ EBAQ= = )) )
ni tip; B = 900 gc AQE = 900 gcEQF= 900
Tng t gc FDP = gc FAP = 450
T gic FDAP ni tip gc D = 900 gc APF = 900 gc EPF = 900 .0,25
Cc im Q, P,C lun nhn EF di 1gc900 nn 5 im E, P, Q, F, C cng nm
trn 1 ng trn ng knh EF 0,25
b. Ta c gc APQ + gc QPE = 1800 (2 gc k b) gc APQ = gc AFE
Gc AFE + gc EPQ = 1800
Tam gic APQ ng dng vi tam gic AEF (g.g)
2
2 1 1 222APQ
APQ AEE
AEF
S k S SS
= = = =
c. gc CPD = gc CMD t gic MPCD ni tip gc MCD = gc CPD (cngchn cung MD)
Li c gc MPD = gc CPD (do BD l trung trc ca AC)
gc MCD = gc MDC (do M thuc trung trc ca DC)
gc CPD = gcMDC = gc CMD = gcMCD tam gic MDC u gc CMD =600
tam gic DMA cn ti D (v AD = DC = DM)V gc ADM =gcADC gcMDC = 900 600 = 300
gc MAD = gc AMD (1800 - 300) : 2 = 750 gcMAB = 900 750 = 150
Bi 5 t x = 1/a; y =1/b; z = 1/c x + y + z = 0 (v 1/a = 1/b + 1/c = 0)
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x = -(y + z) x3 + y3 + z3 3 xyz = -(y + z)3 + y3 3xyz -( y3 + 3y2 z +3 y2z2 + z3) + y3 + z3 3xyz = - 3yz(y + z + x) = - 3yz .0 = 0T x3 + y3 + z3 3xyz = 0 x3 + y3 + z3 = 3xyz 1/ a3 + 1/ b3 + 1/ c3 3 1/ a3 .1/ b3 .1/ c3 = 3/abcDo P = ab/c2 + bc/a2 + ac/b2 = abc (1/a3 + 1/b3+ 1/c3) = abc.3/abc = 3
nu 1/a + 1/b + 1/c =o th P = ab/c2 + bc/a2 + ac/b2 = 3
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19
Bi 1Cho biu thc A =2
222 12)3(
x
xx ++ 22 8)2( xx +
a. Rt gn biu thc Ab. Tm nhng gi tr nguyn ca x sao cho biu thc A cng c gi tr nguyn.
Bi 2: (2 im)Cho cc ng thng:
y = x-2 (d1)y = 2x 4 (d2)y = mx + (m+2) (d3)
a. Tm im c nh m ng thng (d3 ) lun i qua vi mi gi tr ca m.b. Tm m ba ng thng (d1); (d2); (d3) ng quy .Bi 3: Cho phng trnh x2 - 2(m-1)x + m - 3 = 0 (1)
a. Chng minh phng trnh lun c 2 nghim phn bit.b. Tm mt h thc lin h gia hai nghim ca phng trnh (1) m khng
ph thuc vo m.c. Tm gi tr nh nht ca P = x21 + x2
2 (vi x1, x2 l nghim ca phng trnh(1))Bi 4: Cho ng trn (o) vi dy BC c nh v mt im A thay i v tr trn cungln BC sao cho AC>AB v AC > BC . Gi D l im chnh gia ca cung nh BC. Cctip tuyn ca (O) ti D v C ct nhau ti E. Gi P, Q ln lt l giao im ca cc cpng thng AB vi CD; AD v CE.
a. Chng minh rng DE// BCb. Chng minh t gic PACQ ni tipc. Gi giao im ca cc dy AD v BC l F
Chng minh h thc: CE1
= CQ1
+ CE1
Bi 5: Cho cc s dng a, b, c Chng minh rng: 21
=
=
2
1
y
x
Vy N(-1; 2) l im c nh m (d3) i quab. Gi M l giao im (d1) v (d2) . Ta M l nghim ca h
=
=
42
2
xy
xy=>
=
=
0
2
y
x
Vy M (2; 0) .
Nu (d3) i qua M(2,0) th M(2,0) l nghim (d3)Ta c : 0 = 2m + (m+2) => m= -
3
2
Vy m = -3
2th (d1); (d2); (d3) ng quy
Bi 3: a.'
= m2 3m + 4 = (m -2
3)2 +
4
7>0 m.
Vy phng trnh c 2 nghim phn bit
b. Theo Vit:
=
=+
3
)1(2
21
21
mxx
mxx=>
=
=+
622
22
21
21
mxx
mxx
x1+ x2 2x1x2 4 = 0 khng ph thuc vo ma. P = x1
2 + x12 = (x1 + x2)
2 - 2x1x2 = 4(m - 1)2 2 (m-3)
= (2m -2
5)2 + m
4
15
4
15
VyPmin = 415
vi m =4
5
Bi 4: V hnh ng vit gi thit kt lun
a. S CDE =2
1S DC =
2
1S BD = BCD
=> DE// BC (2 gc v tr so le)
b. APC = 21 s (AC - DC) = AQC
=> APQC ni tip (v APC = AQCcng nhn oan AC)c.T gic APQC ni tip CPQ = CAQ (cng chn cung CQ) CAQ = CDE (cng chn cung DC)Suy ra CPQ = CDE => DE// PQ
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Ta c:PQ
DE=
CQ
CE(v DE//PQ) (1)
FC
DE=
QC
QE(v DE// BC) (2)
Cng (1) v (2) : 1==+
=+CQ
CQ
CQ
QECE
FC
DE
PQ
DE
=>DEFCPQ
111=+ (3)
ED = EC (t/c tip tuyn) t (1) suy ra PQ = CQ
Thay vo (3) :CECFCQ
111=+
Bi 5:Ta c:cba
a
++
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