20 ĐỀ ÔN THI VÀO LỚP 10 - ÔN THI VÀO THPT - NĂM 2009-2010

8/14/2019 20 N THI VO LP 10 - N THI VO THPT - NM 2009-2010

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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


14/54



)(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


17/54



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


19/54



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


20/54



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+ + = + + = + + =


21/54



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 = = =


22/54



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


23/54



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


24/54



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


25/54



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:


27/54



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


28/54



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


29/54



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


30/54



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


31/54



=

+

+

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


32/54



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 :


33/54



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


34/54



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.


35/54



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


36/54



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

=


37/54



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


38/54



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


39/54



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


40/54



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


41/54



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


42/54



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)


43/54



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


44/54



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


45/54



( )

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)


46/54



(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

+=


47/54



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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20 ĐỀ ÔN THI VÀO LỚP 10 - ÔN THI VÀO THPT - NĂM 2009-2010