DethiToanvaolop10tunam1988-2013

1 thi vo lp 10 ca thnh ph h ni*Nm hc :1988-1989 ( thi 10/8/1988 , tg =150)

Bi 1Cho A=

2

2 22 2 4 3:2 2 4 2

x x x xx x x x x

a/ Rt gn A.b/ Tnh gi tr ca A khi |x | = 1

Bi 2Mt chic xe ti i t tnh A n B vi vn tc 40km/h.. Sau 1gi 30 pht, mt chic xe con

cng khi hnh t tnh A i n tnh B vi vn tc 60km/h. Hai xe gp nhau khi chng i cmt na qung ng AB.

Tnh qung ng AB.Bi 3

Cho t gic ABCD ni tip trong mt ng trn v P l trung im ca cung AB khng cha Cv D. Hai dy PC v PD ln lt ct AB ti E v F. Cc dy AD v PC ko di ct nhau ti I: cc dyBC v PD ko di ct nhau ti K. Chng minh rng:

a/ Gc CID bng gc CKD.b/ T gic CDFE ni tip c.c/ IK // AB.d/ ng trn ngoi tip tam gic AFD tip xc vi PA ti A.

Bi 4:Tm gi tr ca x biu thc :M = ( 2x - 1)2 3 |2x-1| + 2

t gi tr nh nht v tm GTNN .

GI GII thi vo THPT 1988-1989Bi I:

1/ k: x 0 ; x 2 & x 3

A =2

2 22 2 4 3:2 2 4 2

x x x xx x x x x

=

22 2 4 3:2 2 (2 )(2 ) (2 )x x x xx x x x x x

` =2 2 2(2 ) (2 ) 4 (2 ).(2 )(2 ) 3

x x x x xx x x

=

2 2 24 4 4 4 4 (2 ).(2 )(2 ) 3x x x x x x x

x x x

=24 8 (2 ).(2 )(2 ) 3x x x xx x x

=4 ( 2) (2 ).(2 )(2 ) 3x x x xx x x

=243

xx

22/ |x| = 1=>4 21 34 11 3

A

A

Bi II:Gi di qung ng AB l x (km ; x > 0)Ta c phng trnh:

3: 40 : 602 2 2x x

Bi III:a/ CID =CKD v l cc gc chn cc cung bng nhau.(=> CDIK ni tip)b/ T gic CDEF ni tip c v gc ngoi bng gc trong khng k vi n.c/ IK//AB v t gic CDIK ni tip => IKD = ICD & ICD = PFB ( t gic CDEF ni

tip) => K lun .d/ AF l tt t(AFD) v EAF = ADF (nt chn cc cung bng nhau).-

Bi IV:

M = ( 2x - 1)2 3 |2x-1| + 2 = (| 2x 1|)2 3 |2x-1| + 94 -14

= ( |2x 1| 32 )2 - 14 -

14

Du = xy ra khi ( |2x 1| 32 )2 = 0 | 2x - 1| = 32

2x 1 = 32 32 1 2

32 1 2

x

x

1

2

54

14

x

x

.............................................................................................................

thi vo lp 10 ca thnh ph h ni*

Nm hc :1989-1990

K

F

E

P

O

D

C

B

A

I

3Bi 1Cho biu thc

A = 1- ( 22 5 1

1 2 4 1 1 2x

x x x ) : 21

4 4 1x

x x

a/ Rt gn A v nu cc iu kin phi c ca x.

b/ Tm gi tr ca x A = 12

Bi 2Mt t d nh i t tnh A n tnh B vi vn tc 50km/h. Sau khi i c 2/3 qung ng

vi vn tc , v ng kh i nn ngi li xe phi gim vn tc mi gi 10km trn qung ng cnli. Do t n tnh B chm hn 30 pht so vi d nh. Tnh qung ng AB.Bi 3

Cho hnh vung ABCD v mt im E bt k trn cnh BC. Tia Ax vung gc vi AE ct cnhCD ko di ti F. K trung tuyn AI ca tam gic AEF v ko di ct cnh CD ti K.ng thng quaE v song song vi AB ct AI ti G.

a/ Chng minh AE = AF.b/Chng minh t gic EGFK l hnh thoi.c/ Chng minh tam gic AKF v CAF ng dng v AF2 = KF.CFd/Gi s E chuyn ng trn cnh BC, chng minh rng FK = BE + DK v chu vi tam gic ECK

khng i.Bi 4

Tm gi tr ca x biu thc y=2

22 1989x xx

(k x 0) t gi tr nh nht v tm GTNN

.

GI GII 1989-1990Bi I:

A = 1- ( 22 5 1

1 2 4 1 1 2x

x x x ) : 21

4 4 1x

x x

1/k x & x 1A = 1- ( 2 5 11 2 (2 1)(2 1) 2 1

xx x x x ) : 2

1(2 1)xx

4= 1- 2(2 1) 5 2 1(2 1)(2 1)x x xx x .

2(2 1)1

xx = 1-

4 2 5 2 1(2 1)(2 1)x x x

x x .

2(2 1)1

xx

= 1- 1(2 1)(2 1)x

x x

.2(2 1)

1xx = 1-

2 12 1xx =

22 1x

2/ A = - 12 2

2 1x = -

12 2x - 1 = 4 x = 2,5

Bi II:Gi qung ng AB l x (km & x >0 )Ta c phng trnh

2 1 1: 50 : 403 3 50 2xx x 2 1150 120 50 2

x x x

Bi III:a/ AE = AF. V FAD = EAB (cng ph viDAE)

=> ADB = ABE (cnh gv- gn ) => k lun.b/ Cc tam gic vung IGE & IKF bng nhau (GE // KT

IE = IF) => GF = GE =KF = KE (v AK l trung trc).c/ tam gic AKF v CAF ng dng v AF2 = KF.CF

V ABCD l hnh vung => goc ACF = 450V tam gic AEF vung cn &AI l trung trc goc FAK = 450 => 2 tam gic ng dng (gg). T s => k lun

d/ FD = BE (V 2 tam gic bng nhau) => FK = BE+DK CECK = FK + KC + EC & CD DK = CK = BE ; CE = DK CECK = 2BC (khng i).

Bi IV: y =2

22 1989x xx

(k x 0 => y 0 ) t gi tr nh nht 1y t gi tr ln nht

2

2 2 1989x

x x max 2

12 19891 x x

max 22 19891 x x min

M 22 19891 x x = 2 2

1989 2 1989.(1988 1)1989x x

= 1989 ( 2 21 1 1 12. .1989 1989x x ) +19881989

G

K

I

F

E

D

C

B

A

5= 1989. ( 1 11989x )2 + 19881989

19881989 => Min y =

19891988 khi x = 1989.

thi vo lp 10 ca thnh ph h ni

Nm hc :1990-1991

Bi 1:Xt biu thc

P = ( 1 1 59 13 1 3 1x x

xx x ) : (1-

3 23 1

xx )

a/ Rt gn P.

b/ Tm cc gi tr ca x P = 65Bi 2

Mt xe ti v mt xe con cng khi hnh t tnh A n tnh B. Xe i vi vn tc 30km/h, xe coni vi vn tc 45km/h. Sau khi i c qung ng AB, xe con tng vn tc thm 5km/h trn qungng cn li. Tnh qung ng AB, bit rng xe con n tnh B sm hn xe ti 2 gi 20 pht.Bi 3:

Cho ng trn (O), mt dy AB v mt im C ngoi trn nm trn tia AB. T im chnhgia ca cung ln AB k ng knh PQ ca ng trn , ct dy AB ti D.Tia CP ct ng trn tiim th hai I.Cc dy AB v QI ct nhau ti K.

a/ Cm t gic PDKI ni tip c.b/ Cm CI.CP = CK.CDc/ Cm IC l tia phn gic ca gc ngoi nh I ca tam gic AIBd/ Gi s A,B,C c nh. Cmr khi ng trn (O)thay i nhng vn i qua B th ng thng

QI lun i qua mt im c nh.Bi 4

Tm gi tr ca x biu thcy = x - 1991x t gi tr nh nht v tm GTNN .

GI GII 1990-1991Bi I:

61/ k: x 1/9 => P = ( 1 1 59 13 1 3 1x x

xx x ) : ( 1-

3 23 1

xx )

= ( 1)(3 1) (3 1) 5(3 1)(3 1)x x x x

x x

:3 1 3 2

3 1x x

x

= 3 3 1 3 1 5(3 1)(3 1)x x x x x

x x

.3 1

3x = 3(3 1)(3 1)

xx x .

3 13x = 3 1

xx

2/ P = 65 3 1xx =

65 => 5x 6 (3 1x ) = 0 5x - 18 x +6 = 0

= => x =Bi II:

Gi qung ng AB l x(km, x > 0)

Ta c phng trnh: 3 1 1. . 230 4 45 4 50 3x x x

Bi IIIa/ t gic PDKI ni tip c v PDK = PIK = 900b/ CI.CP = CK.CD v ICK ~ DCPc/ IC l tia pg v IQ l pg AIB v IC IQd/ K l im c nh v IC, IK l cc phn gic trong v ngoiti I ca tam gic AIB ( chia iu ha)KB IB CBKA IA CA m A,B,C c nh.

Bi IV:Tm gi tr ca x biu thc

y = x - 1991x t gi tr nh nht

y = x - 1991x = [( x 1991)- 1991x + 14 ] -14 + 1991

= ( 1991x - 12 )2 + 31990 4

14 +

31990 4 = 1991 => Min y = 1991 khi x = 1991

...............................................................................................................................


Nm hc :1991-1992

K

D

I

O

Q

P

C

B

A

7Bi 1Cho biu thc

Q= ( 3 19x xx ) : (

9 3 2( 3)( 2) 2 3

x x xx x x x

)

a/ Rt gn Q.b/ Tm gi tr ca x Q < 1

Bi 2 Mt on xe vn ti d nh iu mt s xe cng loi i vn chuyn 40 tn hng. Lc sp khihnh , on xe c giao thm 14 tn na. Do , phi iu thm 2 xe cng loi trn v mi xe phich thm 0,5 tn. Tnh s lng xe phi iu theo d nh. Bit rng mi xe ch s hng nh nhau.Bi 3

Cho on thng AB v mt im C nm gia A,B. Ngi ta k trn na mt phng b AB hai tiaAx v By vung gc vi AB v trn tia Ax ly mt im I. Tia vung gc vi CI ti C ct tia By ti K.ng trn ng knh IC ct IK ti P.

a/ Cm t gic CPKB ni tip c .b/ Cm AI.BK= AC.CBc/ Cm tam gic APB vungd/ Gi sA,B,I c nh. Hy xc nh v tr ca im C sao cho din tch hnh thang vung ABKI

ln nht.Bi 4

Chng minh rng cc ng thng c phng trnh y = (m-1)x + 6m - 1991 (m ty )lun iqua mt im duy nht m ta c th xc nh c ta ca n.

GI GII thi vo lp 10 ca thnh ph h niNm hc :1991-1992

Bi I:a/k: x 0 , x 4 & x 9

=> Q = ( 3 19x xx ) : (

9 3 2( 3)( 2) 2 3

x x xx x x x

)

= 3 9( 3)( 3)x x xx x :

9 ( 3)( 3) ( 2)( 2)( 3)( 2)

x x x x xx x

8= 3( 3)( 3)( 3)x

x x :

9 9 4( 3)( 2)x x xx x

=

3( 3)x .

( 3)( 2)( 2)( 2)x xx x

=3

2x

b/ Tm gi tr ca x Q < 1 3 2x < 1 2x > 3 x > 1 x >1 (x 4 & x 9)

Bi II:Gi s xe d nh iu l x ( x (~ N* )Ta c phng trnh

40 40 14 12 2x x

Bi III:a/ t gic CPKB ni tip c v CPK = CBK = 900b/ AI.BK= AC.CB v AIC ~ BCK (gg)c/ APB vung v APB = APC + BPCm APC = AIC = KGB, BPC = BKC => KLd/ SABKI = AB.(AI + BK)

-Bi IV:y= (m-1)x + 6m - 1991 = mx x + 6m - 1991= m (x + 6) 1991 => Nu x = - 6 th y = - 1991 + 6 = - 1985

Vy ta c A (-6 ; - 1985) c nh.

.. thi vo lp 10 ca thnh ph h ni*

Nm hc :1992-1993

Bi 1:Cho biu thc

B = ( 2 11 1x x

x x x ) : (1-

21

xx x

)

a/ Rt gn B.b/ Tm B khi x = 5+ 2 3Bi 2:

O

P

K

I

C

B

A

9Hai ngi th cng lm mt cng vic trong 7 gi 12 pht th xong. Nu ngi th nht lmtrong 5 gi, ngi th 2 lm trong 6 gi th c hai ngi lm c cng vic. Hi mi ngi lmmt mnh cng vic th my gi xong.Bi 3:

Cho na ng trn ng knh AB. K l im chnh gia ca cung AB. Trn cung KB ly M(M K,B ). Trn tia AM ly N sao cho AN = BM. K dy BP//KM. Gi Q l giao im ca cc ngthng AP, BM.

a/ So snh cc tam gic AKN v BKM.b/ Cm tam gic KMN vung cn.c/ T gic ANKP l hnh g? Ti sao?d/ Gi R,S ln lt l giao im th 2 ca QA v QB vi ng trn ngoi tip tam gic OMP,

chng minh khi M di ng trn cung KB th trung im I ca RS lun nm trn ng trn c nh.Bi 4

Gii phng trnh1 2 2

1 21x

x xx

GI GII thi vo lp 10 ca thnh ph h niNm hc :1992-1993

Bi I:

k: x 0 & x 1 => B = ( 2 11 1x x

x x x ) : (1-

21

xx x

)

= 2 1( 1)( 1)x x x xx x x :

1 21

x x xx x

= 1( 1)( 1)x

x x x

.1

1x x

x =

11x

b/ Tm B khi x = 5+ 2 3

B = 15 2 3 1 =1

2(2 3) =2 3

2 => B = 2 32

= 3 12

Bi II:

10

Gi thi gian lm mt mnh xong cng vic ca th nht l x(gi, x > 17 5 )

Thi gain ngi th hai lm mt mnh xong cng vic l y (gi, y > 17 5 )

Th trong 1 gi, ngi th nht lm c 1x (cv); ngi th hai lm c1y (cv) & c hai lm c

536 (cv). => ta c h phng trnh:

1 1 536

5 6 34

x y

x y

Bi III:a/tam gic AKN = BKM. (cgc)b/ tam gic KMN vung cn v KN = KM (2 tgbn)& AKN + NKB = NKB + MKBc/ T gic ANKP l hnh bh v PAN = KMN

= KNM = 450& RPK = APK (tgnt) = PAN = 450

d/ ABM = RPM (ABMP nt) RPM = QSR (RPMS nt) => RS//AB

BP//KM => cung KP = cung MB => POM = 900=> OMP ni tip ng trn ng knh PM (k i)=> Q = 450 (k i)K IE // AQ , IF // BQ => EIF = 450 khng i, RS = OM = OB = OA k i =>E, F l trung imca OA v OB => E, F c nh=> E(~ cung 450 v trn on EFBi IV:Gii phng trnh

1 2 21 21

xx xx

P

F

E

S

R

N

M

I

K

O

B

A

Q

11

........................................................................................................................................................... thi vo lp 10 ca thnh ph h ni

Nm hc :1993-1994

Bi 1:Cho biu thc

M = 1 2 1 2( 1) : (1 )2 1 2 1 2 1 2 1x x x x x xx x x x

a/ Rt gn M

b/ Tnh M khi x = 12 (3+2 2 )

Bi 2:Hai vi nc cng chy vo mt b khng c nc v chy y b trong 4 gi 48 pht. Nu chy

ring th vi th nht c th chy y b nhanh hn vi th hai 1 gi.Hi nu chy ring th mi vi schy y b trong bao lu?Bi 3:

Cho 2 ng trn (O 1 ) v ( O 2 ) tip xc ngoi nhau ti A v tip tuyn chung Ax. Mt ngthng d tip xc vi (O 1 ) , ( O 2 ) ln lt ti cc im B,C v ct Ax ti M.K cc ng knh B O 1 D,C O 2 E.

a/ Cmr M l trung im ca BC.b/ Cmr tam gic O1MO2 vung.c/ Cmr B,A,E thng hng; C,A,D thng hng.

d/ Gi I l trung im ca DE. Cmr ng trn ngoi tip tam gic IO1O2 tip xc vi ng thng BC.Bi 4:Tm m h phng trnh sau y c nghim

x2- (2m-3)x + 6 = 02 x2 +x + (m-5) =0

HNG DN GII thi vo lp 10 ca thnh ph h niNm hc :1993-1994

Bi 1:a/ Rt gn; k x 0 & x

12

M = 1 2 1 2( 1) : (1 )2 1 2 1 2 1 2 1x x x x x xx x x x

= ( 1)( 2 1) ( 2 )( 2 1) (2 1) 2 1 ( 1)( 2 1) ( 2 )( 2 1):( 2 1)( 2 1) ( 2 1)( 2 1)x x x x x x x x x x x x

x x x x

= 2 2 1 2 2 2 2 1 2 1 2 2 1 2 2 2:( 2 1)( 2 1) ( 2 1)( 2 1)x x x x x x x x x x x x x x x x

x x x x

= 2 2 2 2 2 2:( 2 1)( 2 1) ( 2 1)( 2 1)x x xx x x x

=

2 2 ( 1) ( 2 1)( 2 1).( 2 1)( 2 1) 2( 1)x x x x

x x x

= - 2x

b/ Tnh M khi x = 12 (3+2 2 ) =12 ( 2 + 1)

2

M = - 2( 2 1) = - ( 2 + 1)Bi 2:

Gi thi gian vi I chy mt mnh y b l x (h, x > 4 45 )

Thi gian vi II chy mt mnh y b l y (h, y > 4 45 )

Th trong 1h vi I chy c 1x (b), vi II chy c1y (b) & c hai vi chy c 1 : 4

45 (b)

Ta c h phng trnh

1 1 5 1 24 x y 1 2x y

Bi 3:

a/ Cm M l trung im ca BC.MA MBMB MC

=> MB = MC (t/c 2 tt ct nhau) => Kl

b/ Cm O1MO2 vung.V MA = MB = MC (cmt) => ABC vung ti AM 1ABM AO M (gnt, gc tm)V 2ACM AO M = > 1 2AOM AO M = 900 => KL

c/ Cm B,A,E thng hng; C,A,D thng hng.

I

A

E D

M

C

B

O1 O2

13

V ABC vung ti A(cmt) => BAC = 900 & EAC = 900 (gnt chn na ng trn) => KLTng t vi C , A, D.

d/ Cm BC l tt t(IO1O2) ADE vung ti A(do ) = >ID = IA = IE (t/c) => O1I l trung trc ca AD => O1I // O2M, tng tta c O2I // O1M m 1 2O MO = 900 => t gic O1MO2I l hnh ch nht => tm t ngoi tip IO1O2l giao im 2 cho IM v O1O2. T gic BCED l hnh thang vung ( B = 900) => IM l ng trungbnh => IM BC => BC l tt t(IO1O2).(C th dng t/c ng trung bnh ca tam gic cm t gic O1MO2I l hnh bnh hnh & 1 2O MO =900

=> t gic O1MO2I l hnh ch nht ).


Nm hc :1994-1995

Bi 1: Cho biu thc P =3

32 1 1.1 11a a a aa a aa

a) Rt gn P

b) Xt du ca biu thc P. a1Bi 2: Gii bi ton bng cch lp phng trnh

Mt ca n xui t A n B vi vn tc 30km/h, sau li ngc t B v A. Thi gian xui t hn thigian ngc 1h20 pht. Tnh khong cch gia hai bn A v B bit rng vn tc dng nc l 5km/h vvn tc ring ca ca n khi xui v ngc l bng nhau.Bi 3:

Cho tam gac ABC cn ti A, A < 900, mt cung trn BC nm trong tam gic ABC v tip xc viAB,AC ti B v C. Trn cung BC ly mt im M ri h ng vung gc MI,MH,MK xung cc cnhtng ng BC ,CA, BA. Gi P l giao im ca MB,IK v Q l giao im ca MC,IH.

a) Chng minh rng cc t gic BIMK,CIMH ni tip cb) Chng minh tia i ca tia MI l phn gic ca gc HMKc) Chng minh t gic MPIQ ni tip c. Suy ra PQ//BCd) Gi (O1) l ng trn i qua M,P,K,(O2) l ng trn i qua M,Q,H; N l giao im th hai ca

(O1) v (O2) v D l trung im ca BC. Chng minh M,N,D thng hng.

14

Bi 4: Tm tt c cc cp s (x;y) tho mn phng trnh sau:

5x- 2 01)2( 2 yyx

HDG thi vo lp 10 ca thnh ph h ni*

Nm hc :1994-1995

Bi 1: a/Rg biu thc (k : x 0 & x 1 )

P =3

32 1 1.1 11a a a aa a aa

= 2 1 ( 1) 1( 1)( 1)a a a a a aa a a

= 22 1 1( 1)( 1)a a a aa a a = 21 1( 1)( 1)a a aa a a = 1a c) Xt du ca biu thc P. a1P. a1 = ( 1a ). a1 Vi a 0 v a < 1 th a < 1 => 1a P. a1 < 0.

Bi 2: Gii bi ton bng cch lp phng trnh

Gi khong cch gia 2 bn l x (km; x > 0)

Th thi gian xui l 30x (h). Thi gian ngc l 20

x (h)

Ta c phng trnh 20x - 30

x = 43Bi 3:

a/Chng minh cc t gic BIMK,CIMH ni tip c

MK AB (gt) => MKB = 900 & MI BC (gt)=> MIB = 900 BIMK ni tip c

Tng t vi t gic CIMHb/ C/m tia i ca tia MI l phn gic ca HMKGi tia i ca MI l Mx, ta c:V t gic BIMK ni tip (cmt) => xMK = IBK (cng b KMI )V t gic CIMH ni tip (cmt) => xMH = ICHM IBK = ICH (cng chn cung BC) => xMK = xMH => KLc/Chng minh t gic MPIQ ni tip c. Suy ra PQ//BC

PMQ = s cung ln BCPIM = KBM (nt chn cung KM) = s cung BM

x

Q

P

K

H

C

B

I

M

A

15

QIM = HCM (nt chn cung HM) = s cung MC PMQ + PIM + QIM = 1800 => t gic MPIQ ni tip c=> PQM = PIM , PIM = KBM & KBM = ICM PQM = ICM => PQ//BC

thi tt nghip thcs thnh ph h ni*

Nm hc :1995-1996

A/ l thuyt : Hc sinh chn 1 trong 2 1: Pht biu nh ngha v nu cc tnh cht ca hm s bc nht.

Trong 2 hm s sau y, hm s no l hm s b nht ? V sao?

y = 1 2x ; y = x + 1x 2 : Pht biu du hiu nhn bit hnh bnh hnh.B/ Bi tp1/ Xt biu thc

B =( 11aa -

11

aa -

81a

a ) : (3

1a aa -

11a )

a) Rt gn B.b) So snh B vi 1.2/ Gii bi ton bng cch lp phng trnhNu hai vi nc cng chy vo mt b , th sau 6 gi y. Nu vi 1 chy 20 pht v vi 2 chy

30 pht th c 16 b.

Hi nu mi vi chy mt mnh th phi bao lu mi y b ?Bi 3Cho na ng trn ng knh AB v 2 im C,D thuc na dng trn sao cho cung AC < 900 v

gc COD = 900. Gi M l mt im trn na ng trn, sao cho C l im chnh gia cung AM. Ccdy AM v BM ct OC, OD ln lt ti E, F.a/ T gic OEMF l hnh g? Ti sao?

16

b/ Chng minh D l im chnh gia cung MB.c/ ng thng d tip xc vi na ng trn ti M v ct cc tia OC, OD ln lt ti I v K. Chngminh rng t gic OBKM v OAIM ni tip c.

GI GII tn 1995-1996Bi I:

a/ B = 4 4a

a

b/ Xt bt B -1 = 4 4a

a - 1=2( 2) 04

aa

=> B = 1 khi a = 4.

Bi II:

H pt:1 1 1

61 1 13 2 15

x y

x y

1015

xy

Tg vi 1 chy = 10h, tg vi 2 chy = 15h.Bi III:

a/ MEOF l hcn v c 3 gc vung.b/ OD MB =>c/ KM & KB l tip tuyn nn gc OMK = gc OBK = 900

thi vo lp 10 ca thnh ph h ni

Nm hc :1995-1996

Bi1: Cho biu thc A =

12

21:11

1aa

aa

aaa) Rt gn Ab) Tm GT ca a A>1/6

Bi2: Cho phng trnh x2-2(m+2)x+m+1=0 (n x)

17

a) Gii phng trnh khi m = - 23

b) Tm cc GT ca m phng trnh c hai nghim trI duc) Gi x1,x2 l hai nghim ca phng trnh .Tm GT ca m

x1(1-2x2)+ x2(1-2x1) =m2

Bi 3: Cho tam gic ABC(AB>AC ; BAC >900). I,K th t l cc trung im ca AB,AC. Cc ngtrn ng knh AB,AC ct nhau ti im th hai D; tia BA ct ng trn (K) ti im th hai E, tiaCA ct ng trn (I) ti im th hai F.

a) Chng minh bai im B,C,D thng hngb) Chng minh t gic BFEC ni tip.c) Chng minh ba ng thng AD,BF,CE ng quyd) Gi H l giao im th hai ca tia DF vi ng trn ngoi tip tam gic AEF. Hy so snh

di cc on thng DH,DE.Bi4: Xt hai phng trnh bc hai : ax2+bx+c = 0; cx2 +bx+a = 0.

Tm h thc gia a,b,c l iu kin cn v hai phng trinh trn c mt nghim chung duy nht.

Gi gii thi vo lp 10 ca thnh ph h ni

Nm hc :1995-1996Bi1: a/ Rg biu thc (k a > 0 & a 1)

A=

12

21:11

1aa

aa

aa

= 1 ( 1)( 1) ( 2)( 2):( 1) ( 2)( 1)a a a a a aa a a a

=1 1 4:( 1) ( 2)( 1)

a aa a a a

= 1 ( 2)( 1). 3( 1)a a

a a

=2

3aa

b/Tm GT ca a A>1/6

16A

23aa > 16

23aa - 16 > 0

2( 2)6

a aa

> 0 2 46a a

a > 0

4a > 0 (v 6 a > 0 ) a > 4 a > 16 (tmk)Bi2: Cho phng trnh x2-2(m+2)x+m+1=0 (n x)

a/Gii phng trnh khi m = - 23

18

Ta c x2 - 2(- 23 +2)x - 2

3 +1= 0 x2 - x - 12 = 0 2x2 2x 1 = 0

= 1 + 2 = 3 =>1

2

1 32

1 32

x

x

b/Tm cc GT ca m phng trnh c hai nghim tri du

1 2

' 0. 0x x

2( 2) ( 1) 01 0

m mm

2 4 4 1 0

1m m mm

2 3 3 0

1m mm

2 3 3 0

1m mm

2 3 9 32 02 4 4

1m mm

23 3( ) 02 41

mm

m < - 1 ( 23 3( ) 02 4m m )

Bi 3:

a/Chng minh bai im B,C,D thng hng

ADB ADC = 900 (gc ni tip chn na ng trn)b/Chng minh t gic BFEC ni tip.

V BFC = BEC = 900 => nt (l)c/Chng minh ba ng thng AD,BF,CE ng quyV AD , BF, CE l cc ng cao ca ABC => ng quy


Nm hc :1996-1997Kha thi ngy 28-29-30/V/1997

A/ L thuyt (2). Hc sinh chn 1 trong 2 :

K

I

E

F

D

C

B

A

19

I: Hy chng minh cng thc

a ab b Vi a 0 v b>0

p dng tnh: 18 1625 50

II: nh nghi ng trn. Chng minh rng ng knh l dy cung ln nht ca ng trn.B. Bi ton bt buc.I. i s (4 im)1)(2) Cho biu thc:

P= 2 4 2 21 1 1a a

a a a a a

a) Rt gn P.b) Tnh gi tr ca P khi a = 3- 2 2

2) (2) Gii bi ton bng cch lp phng trnhMt ngi d nh sn xut 120 sn phm trong mt thi gian nht nh. Do tng nng sut 4 snphm mi gi, nn hon thnh sm hn d nh 1 gi. Hy tnh nng sut d kin ca ngi .II. Hnh hc (4 )Cho ng trn (O;r) v dy cung AB (ABAB. T C khai tip tuyn vi ng trn ti P,K. Gi I l trung im AB.a) Chng minh t gic CPIK ni tip c trong ng trn.b) Chng minh 2 tam gic ACP v PCB l ng dng. T suy ra: CP2 = CB.CAc) Gi H l trc tm ca tam gic CPK. Hy tnh PH theo r.d) Gi s PA// CK, chng minh rng tia i ca tia BK l tia phn gic ca gc CBP

GI GII tn 1996-1997Bi I:

1/ P = 1a

a a

2/ a = 23 2 2 ( 2 1) => P = 2 2 17

Bi II:

20

Gi nng sut d kin l x (sp/h & x nguyn dng)

Pt: 120 120 14x x x1 = 20 (tmk) & x2 = -24 (loi)

Bi III:1/Gc OIC = 900 (I l trung im ca AB)

Gc CPO = gc CKO (tc tip tuyn) => CPIK nt

2/ ACP ~ PCB => CP CACB CP => CP2 = CA.CB

3/ H (~ OC (H l trc tm) => t gic OPHK l hnh thoi => OP = r.4/BKC = BPK (cng chn cung BK ) KBC = BKP (cung AK = cung PK)=> KBC = PKB => Kt lun.


Nm hc :1996-1997( thi 21/7/1996 tg 150)

Bi 1: Cho biu thc

A =

1

21

1:122

11

xxxxxxx

x1) Rt gn A2) Vi GT no ca x th A t GTNN v tm GTNN

Bi 2: Gii bi ton bng cch lp phng trnh

Mt ngi i xe my t A n B cch nhau 120km vi vn tc d nh trc .Sau khi i c 1/3qung ng AB ngi tng vn tc ln 10km/h trn qung ng cn li. Tm vn tc d nh vthi gian ln bnh trn ng,bit rng ngi n B sm hn d nh 24pht.

Bi3:

Cho ng trn (O) bn knh R v mt dy BC c nh. Gi A l im chnh gia ca cung nh BC.Ly im M trn cung nh AC,k tia Bx vung gc vi tia MA I v ct tia CM ti D.

1) Chng minh gc AMD= gc ABC v MA l tia phn giac ca gc BMD.2) Chng minh A l tm ng trn ngoi tip tam gic BCD v gc BDC c ln khng ph

thuc vo v tr im M.

21

3) Tia DA ct tia BC ti E v ct ng trn (O) ti im th hai F, chng minh AB l tip tuyn cang trn ngoai tip tam gic BEF.

4) Chng minh tch P=AE.AF khng i khi M di ng. Tnh P theo bn knh R v ABC =Bi4:

Cho hai bt phng trnh : 3mx -2m>x+1 (1)m-2x 162. Gii bi ton sau bng cch lp phng trnh:Mt t d nh i t tnh A n tnh B vi vn tc 48km/h. Sau khi i mt gi t b chn

ng bi xe ha 10 pht. Do , n tnh B ng hn , xe phi tng vn tc thm 6km/h. Tnhqung ng AB.

3/. Cho ng trn (O;R ), mt dy CD c trung im l H. Trn tia i ca tia DC ly mt imS v qua S k cc tip tuyn SA, SB vi ng trn. ng thng AB ct cc ng thng SO; OH lnlt ti E v F.

a/ Chng minh t gic SEHF ni tip.b/Chng minh OE.OS = R2c/ OH.OF = OE.OS.d/ Khi S di ng trn tia i ca tia DC hy chng minh ng thng AB lun i qua mt im

c nh.

22

GI GII 1997- 1998Bi I:

1/ A = 23aa

2/ A > 16 2

3aa > 16 a > 16

Bi II:Gi qung ng AB l x (km, x > 0).Ta c pt:

48x = 1 + 16 +

4848 6x 120 (tmk)

Bi III:a/T gic SEHF ni tip v SEF = SHF = 900b/ AOS vung ti A => h thc.c/ HOS ~ EOF =>

d/ OH c nh & OF =2R

OH => F c nh.


Nm hc :1997-1998(26/7/1997- tg= 150)

Bi 1Cho biu thc

A = 1 1 2: ( )1 1 1x xx x x x x x

a/Rt gn A.b/ Tm x A = 7

Bi 2:

23

Mt cng nhn d tnh lm 72 sn phm trong mt thi gian nh.Nhng trong thc t xnghip li giao lm 80 sn phm. V vy, mc d ngi lm mi gi thm 1 sn phm song thigian hon thnh cng vic vn tng so vi d nh 12 pht.

Tnh nng sut d kin, bit rng mi gi ngi lm khng qu 20 sn phm.Bi 3:

Cho ng trn O bn knh R, mt dy AB c nh (AB< 2R) v mt im M ty trn cungln AB (M khc A,B). Gi I l trung im ca dy AB v (O) l ng trn qua M v tip xc vi ABti A. ng thng MI ct (O), (O)ln lt ti cc giao im th hai l N,P.

1/ Cm IA2 = IP.IM2/ Cm t gic ANBP l hnh bnh hnh.2/ Cm IB l tip tuyn ca ng trn ngoi tip tam gic MBP.4/ Cm khi M di chuyn th trng tm G ca tam gic PAB chy trn 1 cung trn c nh.

Bi 4:Trong h ta vung gc xOy, cho Parabol y = x2 (P) v ng thng y = x + m (d)Tm m (d) ct hai nhnh ca (P) ti A v B sao cho tam gic AOB vung ti O?

GI GII 1997- 1998Bi I:

1/2/3/

Bi II:1/2/3/

Bi III:--

Bi IV:1/2/3/

24

4/Bi V:

thi tt nghip thcs thnh ph h ni *

Nm hc :1998-1999(C s chn vo lp 10)

A. L thuyt (2 im): Hc sinh chn mt trong hai sau:

1: Pht biu tnh cht c bn ca phn thc i s. Cc ng thc sau ng hay sai,v sao?

35

515255;31

132

2

mm

mm

xx

2: CMR: nu cnh gc vung v cnh huyn ca tam gic vung ny t l vi cnh gc

vung v cnh huyn ca tam gic vung kia th hai tam gic vung ng dng.B. Bt buc(8 im):

Bi1(2,5 im): Cho biu thc P=

141:1

11

123 xx

xxx

x

a) Rt gn Pb) Tm GT nguyn ca x P nhn GT nguyn dng.

Bai 2(2 im): Gii bi ton bng cch lp phng trnh

Mt ngi d nh i xe p t A n B cch nhau 36km trong thi gian nht nh.Sau khi ic na qung ng ngi dng li ngh 18 pht.Do n B ng hn ngi tng vn tc thm 2km/h trn qung ng cn li. Tnh vn tc ban u v thi gian xe lnbnh trn ng.

Bai3(3,5 im):

Cho tam gic ABC vung ti A,ng cao AH. ng trn ng knh AH ct cc cnhAB,AC ln lt ti E v F.1) Chng minh t gic AEHF l hnh ch nht2) Chng minh: AE.AB = AF.AC

25

3) ng thng qua A vung gc vi EF ct cnh BC ti I. Chng minh I l trung im caBC.

4) Chng minh rng: nu din tch tam giac ABC gp i din tch hnh ch nht AEHF thtam gic ABC vung cn.

GI GII 1998 - 1999Bi I:

1/ P = 3x

x

2/ P = 1 + 3 3x => P (~ N khi 3x l c dng ca 3 => x = 16 v x = 36

Bi II:Gi x l vn tc ban u ( x>0 v km/h)Ta c phng trnh :

18 18 3 362 10x x x x1 = 10 (tmk); x2 = -12 (loi)

Bi III:1/ AEH = AFH = A = 900

` 2/ AE.AB = AF.AC = R23/ AEF = C = KAF => IAC cn =>IA = ICTng t, IA = IB => kl

4/ GT => SABC = 4SAFE => t s ng dng k = 2 => EF = CB = AH=> AH = AI => H I => kl

thi tt nghip thcs thnh ph h ni *

Nm hc :1999-2000

A.L thut (2 im): Hc sinh chn mt trong hai sau:

1: Pht biu hai quy tc i du ca phn thc. Vit cng thc minh ho cho tong quy tc.

26

p dng: Thc hin php tnh : abba

baa

2222 .

2: Pht biu nh l v gc ni tip ca ng trn . Chng minh nh l trong trng hp

tm O nm trn mt cnh ca gc.B.Bi ton bt buc(8 im):

Bi1(2,5 im): Cho biu thc P =

12

11:11 xxxxx

x

a) Rt gn Pb) Tm cc GT ca x P>0

c) Tm cc s m c cc GT ca x tho mn P. xmx .Bi 2(2 im): Gii bi ton bng cch lp phng trnh

Mt xe ti v mt xe con cng khi hnh t A i n B.Xe ti i vi vn tc 40km/h, xe coni vi vn tc 60km/h. Saukhi mi xe i c na ng th xe con ngh 40 pht ri chy tpn B; xe ti trn qung ng cn li tng vn tc thm 10km/h nhng vn n B chmhn xe con na gi. Hy tnh qung ng AB.

Bi 3(3,5 im):

Cho ng trn (O) v mt im A nm ngoi ng trn. T A k hai tip tuyn AB,AC vct tuyn AMN vi ng trn( B,C,M,N thuc ng trn; AM 13/ P. xmx x + x - 1- m = 0

k: m > - 1 & m 1Bi II:

Gi qung ng AB l x (km & x > 0)

27

Phng trnh2 1

80 100 60 3 2x x x x = 200 (tmk)

Bi III:1/OE MN v OC AC2/ chng minh BOA = AOC v AOC = BIC3/ chng minh AEC = AOC & AEC = BIC4/SAIN ln nht khi SABN ln nht

SABN ln nht khi B,O,N thng hng.


Nm hc :2000-2001


1: Th no l php kh mu ca biu thc ly cn. Vit cng thc tng qut.

Ap dng tnh : 231

232 .

2: Pht biu v chng minh nh l gc c nh bn trong ng trn.

B.Bi ton bt buc( 8im):

Bi 1(2,5 im): Cho biu thc

P =

22:2

32

4xx

xx

xxxx .

a) Rt gn P

b) Tnh GT ca P bit x= 6-2 5c) Tm cc GT ca n c x tho mn P.( nxx )1 .

Bi 2(2 im): Gii bi ton bng cch lp phng trnh

Mt ca n chy trn sng trong 8h, xui dng 81 km v ngc dng 105km. Mt ln khccng chy trn khc sng ,ca n ny chay trong 4h, xui dng 54km v ngc dng 42km.Hy tnh vn tc khi xui dng v ngc dng ca ca n, bit vn tc dng nc v vn tcring ca ca n khng i.

28

Bai3(3,5 im):

Cho ng trn (O) ng knh AB=2R, dy MN vung gc vi dy AB ti I sao cho IA< IB.Trn on MI ly im E( E khc M v I).Tia AE ct ng trn ti im th hai K.a) Chng minh t gic IEKB ni tip.b) C/m tam gic AME,AKM ng dng v AM2 =AE.AKc) C/m: AE.AK+BI.BA=4R2

d) Xc nh v tr im I sao cho chu vi tam gic MIO t GTLN.

GI GII 2000- 2001Bi I:

1/ P = 1 x2/ x= 6-2 5 = ( 5 -1)2 => P = 2 - 53/ P.( nxx )1 (1 x )( 1x ) > x n

1- x > x n x + x - 1 < - n 1 1 54 4 4x x n ( v k x > 0 & x 4)

21 1 5

4 2 4x n

=> n < 1

Bi II:Gi x l vt xui, y l vt ngc (km/h & x > y > 0).Ta c h phng trnh

81 105 8

54 42 4x y

x y

2721

xy

(tmk)

Bi III:1/ EIB = EKB = 900 => ni tip2/ MAE = KAM

AME = AKM => MAE ~ AKM (gg) => KL3/ AE.AK = AM2

` BI.BA = BM2 ( h thc) => AM2 + BM2 = AB2 = 4R24/CMIO ln nhtMI + IO ln nht

29

Ta c : (MI + IO)2 2(MI2 + IO2) = 2R2

==> chu vi MIO ln nht khi IO = MI = 22R


Nm hc :2001-2002


1: Pht biu nh ngha v nu tnh cht ca hm s bc nht.

Ap dng: Cho hai hm s bc nht y = 0,2x-7 v y = 5-6xHi hm s no ng bin , hm s no nghch bin ,v sao? 2: Nu cc du hiu nhn bit t gic ni tip ng trn.

B.Bi tp bt buc(8 im):

Bi 1(2,5 im): Cho biu thc P =

x

xxx

xxx 1

41:1

2

a) Rt gn Pb) Tm cc GT ca x P

30

GI GII Bi I:

1/2/3/

Bi II:1/2/3/

Bi III:--

Bi IV:1/2/3/4/

Bi V:*


Nm hc :2002-2003(30/5/2003)

A- L thuyt (2) th sinh chn mt trong 2 sau 1, Pht biu v vit dng tng qut ca qui tc khai phng mt tch.

p dng tnh: P = 50 82 .

2. nh ngha ng trn. Chng minh rng ng knh l dy ln nht ca ng trn.B- Bi tp bt buc (8 im)Bi 1 (2,5 )

31

Cho biu thc P = 4 8 1 2( ) : ( )42 2x x x

xx x x x

a/ Rt gn P.b/ Tm gi tr ca x P = -1.c/ Tm m vi mi gi tr ca x>9 ta c:

m( x -3)P >x+1Bi 2 (2). Gii bi ton bng cch lp phng trnh:

Theo k hoch, hai t sn xut 600 sn phm trong mt thi gian nht nh. Do p dng kthut mi nn t I vt mc 18%, t II vt mc 21% , v vy trong thi gian quy nh h hon thnh vt mc 120 sn phm. Hi s sn phm c giao ca mi t theo k hoch?Bi 3 (3,5).

Cho ng trn (O), mt ng knh AB c nh, mt im I nm gi A v O sao cho AI =23 AO. K dy MN vung gc vi AB ti I. Gi C l im ty thuc cung ln MN, sao cho

C khng trng vi M,N v B. Ni AC ct MN ti E.a/ Chng minh t gic IECB ni tip c trong ng trn.b/ Chng minh AME ng dng vi ACM v AM2 = AE.ACc/ Chng minh AE.AC AI.IB = AI2d/ Hy xc nh v tr ca im C sao cho khong cch t N n tm ng trn ngoi

tip tam gic CME l nh nht.

GI GII Bi I:

1/2/3/

Bi II:1/2/3/

Bi III:-

32

-Bi IV:

1/2/3/4/

Bi V:*

..


Nm hc :2003-2004

A-L thuyt(2 im). Th sinh chn mt trong hai sau: 1. nh ngha phng trnh bc nht hai n s v nghim ca n. Hy tm nghim chung ca 2phng trnh : x+ 4y = 3 v x 3y = -4. 2. Pht biu nh l gc c nh bn ngoa ng trn. Chng minh nh l trong trng hp haicnh ca gc ct ng trn.B- Bi tp bt buc (8 im)

Bi 1: Cho biu thc P =

xx

xx

xxx

11:1

a) Rt gn P

b) Tnh GT ca P khi x = 322

c) Tm cc GT ca x tho mn P. 436 xxxBi 2: Gii bi ton bng cch lp phng trnh

hon thnh mt cng vic , hai t phi lm chung trong 6h. Sau 2h lm chung th t hai biu i lm vic khc , t mt hon thnh nt cng vic cn li trong 10h. Hi nu mi tlm ring th sau bao lu s hon thnh cng vic.

Bi3:

33

Cho ng trn (O;R) , ng thng d khng qua O ct ng trn ti hai im phn bit A,B.T mt im C trn d(C nm ngoi ng trn), k hai tip tuyn CM, CN ti ngtrn(M,N thuc O) . Gi H l trung im ca AB, ng thng OH ct tia CN ti K.1) C/m 4 im C,O,H,N thuc mt ng trn2) C/m : KN.KC=KH.KO3) on thng CO ct (O) ti I, chng minh I cch u CM,CN,MN.4) Mt ng thng i qua O v song song vi MN ct cc tia CM,CN ln lt ti E v F.Xc

nh v tr ca im C trn d sao cho din tch tam gic CEF nh nht.

GI GII Bi I:

1/2/3/

Bi II:1/2/3/

Bi III:--

Bi IV:1/2/3/4/

Bi V:*

thi vo TNTHCS +TS lp 10 thnh ph h ni*

34

Nm hc 2004- 2005

Ngy thi 26/5/2005

A/ L thuyt (2): Hc sinh chn 1 trong 2 1: Nu iu kin A c ngha.

p dng : Vi gi tr no ca x th 2 1x c ngha. 2:Pht biu v chng minh nh l gc c nh bn trong ng trn.B. Bi tp bt buc (8)

Bi 1 (2,5) Cho biu thc P = 1 5 4 2( ) : ( )2 2 2x x x

x x x x x

a/ Rt gon P.

b/ Tnh gi tr ca P khi x = 3 52

c/ Tm m c x tha mn P = mx x - 2mx + 1Bi 2 (2) gii bi ton bng cch lp phng trnh

Theo k hoch, mt cng nhn phi hon thnh 60 sn phm trong mt thi gian nht inh.Nhng do ci tin k thut nn mi gi ngi cng nhn lm thm 2 sn phm . V vy , chngnhng hon thnh k hoch sm hn d nh 30 pht m cn vt mc 3 sn phm.Hi theo khoch , mi gi ngi phi lm bao nhiu snr phm?Bi 3 (3,5 )

Cho tam gic ABC vung ti A. Ly im M ty gia A v B. ng trn ng knh BM ctng thng BC ti im th hai l E. Cc ng thng CM, AE ln lt ct ng trn ti cc imtth 2 l H v K.

a/ Cm t gic AMEC l t gic ni tip.b/ cm gc ACM bng gc KHM.c/ cm cc ng thng BH, EM v AC ng quy.d/Gi s AC 0 , x 1 & x 4

35

P = 1 5 4 2( ) : ( )2 2 2x x x

x x x x x

= 1 5 4 (2 ).( 2)[ ] :[ ]2 ( 2) ( 2) ( 2)x x x x

x x x x x x x

= 5 4 ( 2). 4( 2)x x x xx x

=4( 1). ( 2) 14. ( 2)

x x x xx x

2/ x = 3 52 =

26 2 5 ( 5 1) 5 14 4 2x

=> P = 5 1 5 312 2

3/ P = mx x - 2mx + 1 x - 1 = mx x - 2mx + 1Bi II:

1/2/3/

Bi III:--

Bi IV:1/2/3/4/

Bi V:*

thi vo lp 10 thnh ph h ni*

Nm hc :2006- 2007(thi ngay 16/6/2006 120)

36

Bi 1 (2,5 im)

Cho biu thc P = a 3 a 2 a a 1 1:a 1( a 2)( a 1) a 1 a 1

1/ Rt gn biu thc P

2/ Tm a 1 a 1 1P 8

Bi 2 (2,5 im)

Mt ca n xui dng trn mt khc sng t bn A n bn B di 80 km, sau li ngc dng na im C cch bn B 72 km. Thi gian ca n xui dng t hn thi gian ngc dng l 15 pht. Tnhvn tc ring ca ca n bit vn tc ca dng nc l 4km/h.Bi 3 ( 1 im )

Tm to giao im A v B ca th hai hm s y = 2x + 3 v y = x2.Gi D v C ln lt l hnh chiu vung gc ca A v B trn trc honh. Tnh SABCD.

Bi 4 (3 im)

Cho (O) ng knh AB = 2R, C l trung im ca OA v dy MN vung gc vi OA ti C. Gi K lim tu trn cung nh BM, H l giao im ca AK v MN.a) CMR: BCHK l t gic ni tip.b) Tnh AH . AK theo R.c) Xc nh v tr ca im K (KM + KN + KB) t gi tr ln nht v tnh gi tr ln nht .Bi 5 (1 im)

Cho hai s dng x, y tho mn iu kin: x + y = 2. Chng minh: x2y2(x2+y2) 2.

GI GII Bi I:

1/k a 1 & a 0.

=> P = a 3 a 2 a a 1 1:a 1( a 2)( a 1) a 1 a 1

= ( a 2)( a 1) a ( a 1) a 1 a 1:( a 2)( a 1) ( a 1)( a 1) ( a 1)( a 1) ( a 1)( a 1)

= ( a 1) a 2 a:( a 1) ( a 1) ( a 1)( a 1)

37

=1 ( 1)( 1) 1.1 2 2

a a aa a a

2/ 1 a 1 1P 8

2 1 181a a

a

Bi II:Gi vn tc ring ca ca n l x (km/h, x >4)Ta c phng trnh

80 72 14 4 4x x

Bi III:Gii pt: x2 = 2x + 3 x2 2x 3 = 0 x1 = -1 & x2 = 3 (theo Vi et) => y1 = 1& y2 = 9=> A (-1 ; 1) & B (3 : 9)SABCD = (AD + BC ) (|OD| + |OC| ) : 2 (v t gic ABCD l hnh thang vung)

Bi IV:1/ T gic BCHK c C = K = 900 => nt2/ ACH ~ AKB (gg) => AC AHAK AB => AH.AK = AB.AC = R

2

3/ Cm BMN u => KM + KN + KB = 2KN=> max khi KN max = 2R=> K,O,N thng hng (K l im chnh gia cung BM)=> Max(KM + KN + KB) = 4R

(Bi tp 20 /trang 76 /sch BTT9 tp II)

Bi V:

H

N

M

K

C

B

O

A

38

x2y2(x2+y2) = 12 xy. [2xy.(x2 + y2)] 12 xy.

22 222

x xy y = 12 xy.

22( )2

x y = 2xy

22

2x y = 2 (p dung C si cho 2 s dng v x + y = 2 ).


Nm hc :2007-2008 (20/6/2007 120)Bi 1 ( 2,5 im )

Cho biu thc : P = x 3 6 x 4x 1x 1 x 1 Vi x 0 & x 1

1/ Rt gn biu thc P.

2/ Tm x P < 12 .

Bi 2 ( 2,5 im )

Gii bi ton sau bng cch lp phng trnh:Mt ngi i xe p t A n B cch nhau 24 km. Khi t B tr v A ngi tng vn tc ln 4 km/h

so vi lc i, v vy thi gian v t hn thi gian i 30 pht. Tnh vn tc ca xe p khi i t A n B.Bi 3 ( 1 im )

Cho phng trnh x2 + bx + c = 01/ Gii phng trnh khi b = - 3 v c = 2.2/ Tm b, c phng trnh cho c hai nghim phn bit v tch ca chng bng 1.Bi 4 ( 3,5 im )

Cho ng trn (O; R) tip xc vi ng thng d ti A. Trn d ly im H khng trng vi im Av AH < R. Qua H k ng thng vung gc vi d, ng thng ny ct ng trn ti hai im E v B( E nm gia B v H ).

1/ Chng minh ABE EAH v ABH EAH.2/ Ly im C trn d sao cho H l trung im ca on thng AC, ng thng CE ct AB ti K. Chngminh AHEK l t gic ni tip.

3/ Xc nh v tr im H AB = R 3 .Bi 5 ( 0,5 im )

39

Cho ng thng y = ( m - 1 ) x + 2Tm m khong cch t gc to n ng thng l ln nht.

GI GII 2007-2008Bi I:

1/ P = 11xx

2/ P < 12 11

xx 0)Ta c phng trnh

24 24 14 2x x x = 12

Bi III:

2/ k: gii hpt:2

1 2

0 4 0. 1 1

b cx x c

Bi IV:1/ Hai tam gic ng dng theo trng hp gg2/ HAE = HCE (cgc) => C = HAF , m HAF = B (do 2 tam gic dng)

Mt khc, B + HAB = 900 => C + HAB = 900 => AKE = 900 => AKE + AHE =1800 => nt

3/ H OI AB => AI = AB = 32R => cos ( OAI) = 32 => OAI = 30

0 => BAH=600

=> AH = 32R .

Bi V: th lun i qua A (0;2) c nh khi a = m 1 =0 m =1Gi B l im ct truc honh. K OH AB. Trong tam gic vung OAB ta c:

OH OA. Du = xy ra khi H A m 1 = 0 m = 1

40


Nm hc :2008-2009 (18/6/2008 120)Bi 1 ( 2,5 im )

Cho biu thc: P = 1 x x:x x 1 x x

1/ Rt gn P.2/ Tm gi tr ca P khi x = 4.

3/ Tm x P = 133 .

Bi 2 ( 2,5 im )

Gii bi ton sau bng cch lp phng trnh.Thng th nht hai t sn xut c 900 chi tit my. Thng th hai t I vt mc 15% v t II vt

mc 10% so vi thng th nht, v vy hai t sn xut c 1010 chi tit my. Hi thng th nht mit sn xut c bao nhiu chi tit my.Bi 3 ( 3,5 im )

Cho parabol (P): y = 21 x4 v ng thng (d): y = mx + 1

1/ Chng minh vi mi gi tr ca m ng thng (d) lun ct parabol (P) ti hai im phn bit.2/ Gi A, B l hai giao im ca (d) v (P). Tnh din tch tam gic OAB theo m ( O l gc to ).Bi 4 ( 3,5 im )

Cho ng trn (O) c ng knh AB = 2R v E l im bt k trn ng trn ( E khc A v B ).ng phn gic gc AEB ct on thng AB ti F v ct ng trn (O) ti im th hai l K.1/ Chng minh tam gic KAF ng dng vi tam gic KEA.2/ Gi I l giao im ca ng trung trc on EF vi OE, chng minh ng trn (I) bn knh IE tipxc vi ng trn (O) ti E v tip xc vi ng thng AB ti F.3/ Chng minh MN // AB, trong M v N ln lt l giao im th hai ca AE, BE vi ng trn (I).4/ Tnh gi tr nh nht ca chu vi tam gic KPQ theo R khi E chuyn ng trn ng trn (O), vi P lgiao im ca NF v AK; Q l giao im ca MF v BK.Bi 5 ( 0,5 im )

Tm gi tr nh nht ca biu thc A, bit:A = ( x - 1 )4 + ( x - 3 )4 + 6 ( x - 1 )2 ( x - 3 )2

41

GI GII 2008-2009Bi I:

1/P = 1x xx

2/ P = 7/23/ k x>0 => 3x - 10 x + 3= 0 => x = 9 hoc x = 1/9

Bi II:T I = 400sp; T II = 500spBi III:

1/ => 241 x = mx + 1 24

1 x - mx 1 = 0 => > 0 => ct ti 2 im

2/ SAOB = (| x1| + | x2|) = 2 2 1m Bi IV:

3/ MN l ng knh ca (I) . gc INE = gc OBE (= gc IEN) => MN // AB.4/ Chu vi tam gic KPQ = KP +PQ + KQ = QB + QK + FK = BK + FK BK + FO = ( 2 1)R .Du = xy ra khi E l im chnh gia cung AB.

Bi V:t a = x -2 => A = 8a4 + 8 8Du = xy ra khi x 2 =0 x =2

k thi tuyn sinh vo lp 10 thpt*

Nm hc: 2009-2010(TG=120)

Bi 1 ( 2,5 im )

Cho biu thc : A = x 1 1x 4 x 2 x 2 , vi x 0; x 41/ Rt gn biu thc A.2/ Tnh gi tr ca biu thc A khi x = 25.

3/ Tm gi tr ca x A = -13 .

Bi 2 ( 2,5 im )

42

Gii bi ton sau bng cch lp phng trnh hoc h phng trnh;Hai t sn xut cng may mt loi o. Nu t th nht may trong 3 ngy, t th hai may trong 5 ngy

th c hai t may c 1310 chic o. Bit rng trong mi ngy t th nht may c nhiu hn t thhai 10 chic o. Hi mi t may trong mt ngy c bao nhiu chic o ?Bi 3 ( 1 im )

Cho phng trnh (n x): x2 - 2(m + 1)x + m2 + 2 = 01/ Gii phng trnh cho vi m = 1.2/ Tm gi tr ca m phng trnh cho c hai nghim phn bit x1, x2 tho mn h thc: x12 + x22 =10.Bi 4 ( 3,5 im )

Cho ng trn (O; R) v A l mt im nm bn ngoi ng trn. K cc tip tuyn AB, AC ving trn ( B, C l cc tip im ).1/ Chng minh ABOC l t gic ni tip.2/ Gi E l giao im ca BC v OA. Chng minh BE vung gc vi OA v OE.OA=R2.3/ Trn cung nh BC ca ng trn (O; R) ly im K bt k ( K khc B v C ). Tip tuyn ti K cang trn (O; R) ct AB, AC theo th t ti cc im P v Q. Chng minh tam gic APQ c chu vikhng i khi K chuyn ng trn cung nh BC.4/ ng thng qua O, vung gc vi OA ct cc ng thng AB, AC theo th t ti cc im M, N.Chng minh PM + QN MN.Bi 5 ( 0,5 im )

Gii phng trnh.

2 2 3 21 1 1x x x (2x x 2x 14 4 2 )

GI GII 2009-2010Bi I

1/ A = 2x

x 2/ A=53 3/x =

14

Bi IIT I = 170; T II = 160

Bi III1/ m=1 => x1 =1: x2 =3

43

2/ >0m > x1 + x2 = 10m2 +4m 5 = 0 m1 =1, m2 = -5 => Kt lun m = 1.

Bi IV4/ PMO ~ OQN => PM.QN = OM.ON = MN2 /4

(PM + QN)2 4PM.QN = MN2=> PM + QN MN

Bi V

2 21 1 14 4 2x x x (2x

3 + x2 2x + 1 ) 2 1 1 14 2 2x x (2x + 1)(x2 + 1) K: x -1/2

x + 12 =12 (2x + 1)(x

2 + 1) (2x + 1)x2 = 0 x1 = 0: x2 = -1/2 (Tmk)

k thi tuyn sinh vo lp 10 thpt*

Nm hc: 2010-2011Mn Ton (thi ngy 22/6/2010)

Bi 1(2,5 im):

Cho P = 9&0,993

32

3 xxx

xxx

xx .

1) Rt gn P.

2) Tm gi tr ca x P = 31 .

3) Tm GTLN ca P.

Bi 2(2,5 im): gii bi ton bng cch lp phng trnh

Mt mnh t hnh ch nht c di ng cho l 13m v chiu di ln hn chiu rng l 7m. Tnhchiu di v chiu rng ca mnh t ?

Bi 3(1,0 im):

Cho Parabol (P): y =-x2 v ng thng (d) y =mx-1

44

1) CMR vi mi m th (d) lun ct (P) ti 2 im phn bit.2) Gi x1,x2 l cc honh giao im ca (d) v (P). Tm gi tr ca m x12x2+x22x1- x1x2 =3.

Bi 4(3,5 im):

Cho (O;R) ng knh AB =2R v im C thuc ng trn ( C khc A,B). D thuc dy BC (D khcB,C). Tia AD ct cung nh BC ti E,tia AC ct BE ti F.1) Chng minh t gic FCDE ni tip2) Chngminh DA.DE = DB.DC3) Chng minh CFD = OCB . Gi I l tm ng trn ngoi tip t gic FCDE , chng minh IC l tiptuyn ca (O).4) Cho bit DF =R, chng minh tanAFB = 2.

Bi 5 (0,5 im):

Gii phng trnh x2 +4x +7 =(x+4) 72 x

GI GII 2010-2011Bi I:

1/ A = 3 3x 2/ x = 36 (tmk)3/ MaxA = 1 khi x = 0 (tmk)

Bi II:Gi chiu rng l x, ta c pt: x2 + (x + 7) 2 = 132 => x = 5 => chiu di = 12m.

Bi III:1/ Xt phng trnh: -x2 = mx 1 x2 +mx -1 = 0 , c >0 nn c 2 nghim phn bit => ct

ti 2 im phn bit.2/ Theo nh l Vi et ta c x1 + x2 = -m & x1x2 = - 1 => m = 3.

Bi IV:1/ T gic FCDE ni tip v c 2 gc i bng nhau(=900)2/ADC ~ BDE (gg)

45

3/

4/ Tan AFB = 2 2BC AB RFC DF R (tam gic CBA ~ tam gic CFD )

Bi 5x2 +4x +7 =(x+4) 72 x x2 + 7 - x 2 27 4 7 4 0x x x

2 2 27( 7 ) 4 7 0x x x x x 2 2

2 2 2

2 2

( 7 )( 7 4) 07 0 7 07 4 0 9 3

x x xx x x x xx x x

THI TUYN SINH LP 10 THPT H NI*

Nm hc: 2011 2012 CHNH THC MN: TON

Thi gian lm bi: 120 phtBi I (2,5 im)

Cho x 10 x 5A x 25x 5 x 5 , Vi x 0 v x 25 ta c.

1) Rt gn biu thc A.2) Tm gi tr ca A khi x = 9.3) Tm x A < .

Bi II (2,5 im)Gii bi ton sau bng cch lp phng trnh hoc h phng trnh:Mt i xe theo k hoch ch ht 140 tn hng trong mt s ngy quy nh. Do mi ngy i

ch vt mc 5 tn nn i hon thnh k hoch sm hn thi gian quy nh 1 ngy v ch thmc 10 tn. Hi theo k hoch i xe ch hng ht bao nhiu ngy?Bi III (1,0 im)

Cho parabol (P) : y = x2 v ng thng (d) : y = 2x m2 + 9.1) Tm ta cc giao im ca parabol (P) v ng thng (d) khi m = 1.2) Tm m ng thng (d) ct parabol (P) ti hai im nm v hai pha ca trc tung.

Bi IV (3,5 im)Cho ng trn tm O, ng knh AB = 2R. Gi d1 v d2 ln lt l hai tip tuyn ca ng

trn (O) ti hai im A v B. Gi I l trung im ca OA v E l im thuc ng trn (O) (E khng

46

trng vi A v B). ng thng d i qua im E v vung gc vi EI ct hai ng thng d1, d2 lnlt ti M, N.

1) Chng minh AMEI l t gic ni tip.2) Chng minh gc ENI = gc EBI v gc MIN = 900 .3) Chng minh AM.BN = AI.BI.4) Gi F l im chnh gia ca cung AB khng cha E ca ng trn (O). Hy tnh din tch

ca tam gic MIN theo R khi ba im E, I, F thng hng.Bi V (0,5 im)

Vi x > 0, tm gi tr nh nht ca biu thc: M = 2 14x 3x 20114x

BI GII THI TUYN SINH LP 10 THPT H NI

Nm hc: 2011 2012

Bi I: (2,5 im) Vi x 0 v x 25 ta c :

1) x 10 x 5A x 25x 5 x 5 =( 5) 10 5( 5)

25 25 25x x x xx x x

= 5 10 5 2525 25 25x x x xx x x =

10 2525

x xx

=

2( 5)( 5)( 5)

xx x

= 55xx

2) x = 9 A = 9 5 149 5

3) A < 13 55

xx

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