32 de va dap an ts 10 cac tinh thanh 1415

THY HONG XUN VNH 1

B TUYN SINH 10 TON KHNG CHUYN 14-15

1

S GIO DC V O TO

NINH BNH

THI TUYN SINH LP 10 THPT CHUYN LVT

NM HC 2014 - 2015

Mn thi: TON KHNG CHUYN Ngy thi: 11/6/2014

Thi gian lm bi: 120 pht

( thi gm c 05 cu trong 01 trang)

Cu 1 (3,0 im).

1. Rt gn cc biu thc sau: M 45 245 80

1 1 3 a

N :a 4a 2 a 2

, vi a 0;a 4

2. Gii h phng trnh:x 3y 24

7x y 14

3. Gii phng trnh: 2 2

5x 4 13

x 4x 1 x x 1 3

.

Cu 2 (1,5 im).Trong mt phng ta Oxy, cho Parabol (P): 2y x v ng thng (d): y = mx + 3 (m l

tham s).

a) Khi m = - 2, tm ta giao im ca ng thng (d) v Parabol (P).

b) Tm m ng thng (d) v Parabol (P) ct nhau ti hai im phn bit c honh 1x v 2x tha

mn iu kin: 3 3

1 2x x 10 . Cu 3 (1,5 im). Gii bi ton bng cch lp phng trnh hoc h phng trnh:

Mt phng hp c 440 gh (mi gh mt ch ngi) c xp thnh tng dy, mi dy c s gh bng

nhau. Trong mt bui hp c 529 ngi tham d nn ban t chc phi k thm 3 dy gh v mi dy tng thm

1 gh so vi ban u th va ch ngi. Tnh s dy gh c trong phng hp lc u.

Cu 4 (3,0 im).Cho ng trn tm O ng knh AB. Trn tia tip tuyn Ax ca ng trn ly im M (M

khc A). T M k tip tuyn th hai MC vi ng trn (O). K CH AB ( H AB ). ng thng MB ct (O) ti im Q v ct CH ti im N. Gi I l giao im ca MO v AC. Chng minh rng:

a) T gic AIQM ni tip c. b) OM//BC

c) T s CN

CH khng i khi M di ng trn tia Ax (M khc A).

Cu 5 (1,0 im).Cho a, b, c l cc s thc dng tha mn iu kin a.b.c = 1. Chng minh rng:

3 3 3a b c 3

1 b 1 c 1 c 1 a 1 a 1 b 4

HD Gii bi thy Hong Xun Vnh(THCS Bnh Chiu,Th c):

THY HONG XUN VNH 2


Cu 1:

1. 8024545 M =.=6 5

4

3:

2

1

2

1

a

a

aaN =

a 2 a 2a 2 a 2 a 4 2 a 2

. .33 a 3 aa 2 a 2 a 2 a 2

2.

147

243

yx

yx x 3....

y 7

3. 3

13

1

4

14

522

xx

x

xx

x

V x=0 khng l nghim nn chia t mu v tri pt ban u cho x ta c: 5 4 13

1 1 3x 4 x 1

x x

,v t t=1

xx

ta c: 5 4 13

t 4 t 1 3

3.5 t 1 3.4 t 4 13 t 1 t 4 2 1 223

13t 42t 115 0 t 5hay t13

Vi t=5,suy ra1

xx

=5 25 21

x 5x 1 0 x2

Vi t= 23

13

,suy ra

1x

x =

23

13

213x 23x 13 0 VN

Cu 2:Phng trnh honh giao im (d) v (P): 2x mx 3

a)Khi m=-2,ta c 2 2x 2x 3 x 2x 3 0 x 1v x 3

Suy ra giao im (d) v (P) (1;1);(-3;9)

b)Phng trnh honh giao im (d) v (P): 2x mx 3 2x mx 3 0

2m 12 0 m ,pt lun c 2 nghim pb x1,x2.

p dng Vi-et,ta c 1 2

1 2

S x x m

P x .x 3

Ta c: 10323

1 xx3 3S 3SP 10 m 9m 10 0 m 1

Cu 3:Gi x l s gh trong mt dy ban u (x thuc N)

S dy gh ban u:440

x

Theo bi ta c pt: 440 529

3x x 1

2440 x 1 3x x 1 529x 3x 86x 440 0

K

I

H

N

Q

M

C

x

OBA

THY HONG XUN VNH 3


20x 22(n); x (l)

3 Vy s gh trong phng lc u l 440:22=20 dy.

Cu 4:

a)CM c 0MQA MIA 90 ,suy ra pcm

b)DDCM c do cng vung gc vi AC

c)Gi K l giao im ca BC v Ax.

Do BC// OM,O l trung im AB ,

nn M l trung im AK,

CH//AK do cng vung gc AB,

p dng h qu Ta-let ta c:

CN/KM=NH/AM(=BN/BM).

M KM=AM(M l trung im AK)

nn,CN=NH,suy ra CN/CH=1/2

khng i khi M chy trn Ax.

5.p dng C-si, ta c:

3 3

3a 1 b 1 c a 1 b 1 c 3a

3 . .1 b 1 c 8 8 1 b 1 c 8 8 4

(1)

3 3

3b 1 c 1 a b 1 c 1 a 3b

3 . .1 c 1 a 8 8 1 c 1 a 8 8 4

(2)

3 3

3c 1 b 1 a c 1 b 1 a 3c

3 . .1 b 1 a 8 8 1 b 1 a 8 8 4

(3)

Cng v (1)(2)(3) ta c

3 3 3 3 a b c 6 2 a b ca b c

1 b 1 c 1 c 1 a 1 b 1 a 4 8

3 3 3 a b ca b c 3

1 b 1 c 1 c 1 a 1 b 1 a 2 4

M p dng C-si: 3a b c 3 3. abc

2 2 2

THY HONG XUN VNH 4


Suy ra:

3 3 3a b c 3 3 3

1 b 1 c 1 c 1 a 1 b 1 a 2 4 4

Du= xy ra,khi a=b=c=1

(hoangxuanvinhthuduc.blogspot.com)

2

S GIO DC V O TO H NI MN: TON

K THI TUYN SINH LP 10 THPT Nm hoc: 2014 2015

CHNH THC Thi gian lm bi: 120 pht

Bi I (2,0 im) 1) Tnh gi tr ca biu thc x 1

Ax 1

khi x = 9

2) Cho biu thc x 2 1 x 1

P .x 2 x x 2 x 1

vi x > 0 v x 1

a)Chng minh rng x 1

Px

b)Tm cc gi tr ca x 2P 2 x 5

Bi II (2,0 im) Gii bi ton bng cch lp phng trnh:

Mt phn xng theo k hoch cn phi sn xut 1100 sn phm trong mt s ngy quy nh. Do mi ngy

phn xng sn xut vt mc 5 sn phm nn phn xng hon thnh k hoch sm hn thi gian quy nh

2 ngy. Hi theo k hoch, mi ngy phn xng phi sn xut bao nhiu sn phm?

THY HONG XUN VNH 5


Bi III (2,0 im) 1) Gii h phng trnh:

4 15

x y y 1

1 21

x y y 1

2) Trn mt phng ta Oxy cho ng thng (d): y = -x + 6 v parabol (P): y = x2.

a) Tm ta cc giao im ca (d) v (P).

b) Gi A, B l hai giao im ca (d) v (P). Tnh din tch tam gic OAB.

Bi IV (3,5 im)Cho ng trn (O; R) c ng knh AB c nh. V ng knh MN ca ng trn (O; R)

(M khc A, M khc B). Tip tuyn ca ng trn (O; R) ti B ct cc ng thng AM, AN ln lt ti cc im

Q, P.

1) Chng minh t gic AMBN l hnh ch nht.

2) Chng minh bn im M, N, P, Q cng thuc mt ng trn.

3) Gi E l trung im ca BQ. ng thng vung gc vi OE ti O ct PQ ti im F. Chng minh F l

trung im ca BP v ME // NF.

4) Khi ng knh MN quay quanh tm O v tha mn iu kin bi, xc nh v tr ca ng knh

MN t gic MNPQ c din tch nh nht.

Bi V (0,5 im) Vi a, b, c l cc s dng tha mn iu kin a + b + c = 2. Tm gi tr ln nht ca biu thc

Q 2a bc 2b ca 2c ab

BI GII

Bai I: (2,0 im)

1) Vi x = 9 ta c 3 1

23 1

A

2) a) 2 1 ( 1).( 2) 1

. .( 2) 1 ( 2) 1

x x x x x xP

x x x x x x

1

x

x

b)T cu 2a ta c

2 x 2

2P 2 x 5 2 x 5x

2 x 2 2x 5 x v x > 0

THY HONG XUN VNH 6


2x 3 x 2 0 v x >01

( x 2)( x ) 02

v x >0

1 1x x

2 4

Bai II: (2,0 im)

Gi x l sn phm xng sn xut trong 1 ngy theo k hoch (x > 0)

S ngy theo k hoch l : 1100

x .

S ngy thc t l 1100

x 5. Theo gi thit ca bi ton ta c :

1100

x -

1100

x 5 = 2.

2

1100(x 5) 1100x 2x(x 5)

2x 10x 5500 0

x 50 hay x 55 (loi)

Vy theo k hoch mi ngy phn xng phi sn xut l 50 sn phm.

Bai III: (2,0 im)

1) H phng trnh tng ng vi:

t 1

ux y

v 1

vy 1

. H phng trnh thnh :

4u v 5 8u 2v 10 9u 9 u 1

u 2v 1 u 2v 1 2v u 1 v 1

Do , h cho tng ng :

11

x y 1 x 1x y

1 y 1 1 y 21

y 1

2)

a) Phng trnh honh giao im ca (P) v (d) l

2 6 x x 2 6 0 x x 2hay 3 x x

Ta c y (2)= 4; y(-3) = 9. Vy ta giao im ca (d) v (P) l B(2;4) v A(-3;9)

b) Gi A, B ln lt l hnh chiu ca A v B xung trc honh.

THY HONG XUN VNH 7


Ta c OAB AA'B'B OAA' OBB'S S S S

Ta c AB = B' A' B' A'x x x x 5 , AA = Ay 9 , BB = By 4

Din tch hnh thang : AA'B'BS AA ' BB' 9 4 65

.A 'B' .52 2 2

(vdt)

OAA'S1 27

A 'A.A 'O2 2

(vdt); OBB'S1

B'B.B'O 42

(vdt)

OAB AA'B'B OAA' OBB'

65 27S S S S 4 15

2 2

(vdt)

Bi IV (3,5 im)

1) T gic AMBN c 4 gc vung, v l 4 gc ni tip chn na

ng trn.

2) Ta c ANM ABM (cng chn cung AM)

v ABM AQB (gc c cnh thng gc)

vy ANM AQB nn MNPQ ni tip.

3) OE l ng trung bnh ca tam gic ABQ.

OF // AP nn OF l ng trung bnh ca tam gic ABP

Suy ra F l trung im ca BP.

M AP vung gc vi AQ nn OE vung gc OF.

Xt tam gic vung NPB c F l trung im ca cnh huyn BP.

Xt 2 tam gic NOF = OFB (c-c-c) nn 0ONF 90 .

Tng t ta c 0OME 90 nn ME // NF v cng vung gc vi MN.

4) MNPQ APQ AMN2S 2S 2S 2R.PQ AM.AN 2R.(PB BQ) AM.AN

Tam gic ABP ng dng tam gic QBA suy ra AB BP

QB BA 2AB BP.QB

Nn p dng bt ng thc Cosi ta c 2PB BQ 2 PB.BQ 2 (2R) 4R

Ta c

2 2 2AM AN MNAM.AN

2 2

= 2R2

Do ,2 2

MNPQ2S 2R.4R 2R 6R . Suy ra 2

MNPQS 3R

Du bng xy ra khi AM =AN v PQ = BP hay MN vung gc AB.

A B

P

Q

O

F

E

N

M

THY HONG XUN VNH 8


Bai V: (0,5 im)

Ta c Q 2a bc 2b ca 2c ab

2a bc (a b c)a bc (Do a + b +c = 2)

2 (a b) (a c)a ab bc ca (a b)(a c)2

(p dng BDT vi 2 s dng u=a+b v v=a+c)

Vy ta c 2a bc(a b) (a c)

2

(1)

Tng t ta c : 2b ca(a b) (b c)

2

(2)

2c ab(a c) (b c)

2

(3)

Cng (1) (2) (3) v theo v Q 2(a b c) 4

Khi a = b = c = 2

3th Q = 4 vy gi tr ln nht ca Q l 4.

Trn Quang Hin,Ng Thanh Sn, Nguyn Ph Vinh (THPT Vnh Vin TP.HCM)

3 TRNG THPT CHUYN L QU N BNH NH

---------------- ------------------------------------------------------

CHNH THC Mn thi: TON

Ngy thi: 13/06/2014

Thi gian lm bi: 120 pht (khng k thi gian pht ).

----------------------------------------------------------------------

Bi 1: (2,0 im) Cho biu thc 2 2

11

a a a aA

a a a

, vi a > 0.

a. Rt gn A. b. Tm gi tr ca a A = 2. c. Tm gi tr nh nht ca A.

Bi 2: (2,0 im)

Gi th hm s 2y x l parabol (P), th hm s 4 2 5y m x m l ng thng (d).

a. tm gi tr ca m (d) ct (P) ti hai im phn bit.

b. Khi (d) ct (P) ti hai im phn bit A v B c honh ln lt l 1 2;x x . Tm cc gi tr ca m sao cho 3 3

1 20x x .

Bi 3: (1,5 im )

Tm x, y nguyn sao cho 18x y

THY HONG XUN VNH 9


Bi 4: ( 3,5 im )

Cho ng trn (O) v mt im P ngoi ng trn. K hai tip tuyn PA, PB vi ng trn (O) (A ,B l hai

tip im). PO ct ng trn ti hai im K v I ( K nm gia P v O) v ct AB ti H. Gi D l im i xng ca

B qua O, C l giao im ca PD v ng trn (O).

a. Chng minh t gic BHCP ni tip. b. Chng minh .AC CH

c. ng trn ngoi tip tam gic ACH ct IC ti M. Tia AM ct IB ti Q. Chng minh M l trung im ca AQ. Bi 5: (1,0 im)

Tm gi tr nh nht ca hm s: 2 1

1,y

x x

vi 0< x 0 => A c ngha vi mi 0a .

3

1 2 11

1

a a a aA a a

a a a

b)Tm gi tr ca a A = 2

Ta c: A a a . : A = 2 => 2 2 0a a a a

t: 0a t c pt: 2 2 0t t t1= -1 (loi) t2 = 2 (tha mn iu kin)

Vi t = 2 2 4a a (tha mn iu kin)

Vy: 4a l gi tr cn tm.

c)Tm gi tr nh nht ca A.

Ta c: A a a

2 2 21 1 1 1 1 1

22 2 2 2 4 4

a a a

vi mi a >0

THY HONG XUN VNH 10


( v:

21

02

a

vi mi a > 0)

Du = khi 1 1

02 4

a a (tha mn iu kin 0a )

Vy: 1

4nho nhat

A

khi 1

4a

Bi 2: (2,0 im)

a) Tm gi tr ca m (d) ct (P) ti hai im phn bit

Ta c: (d): 4 2 5y m x m

(P): 2y x

Pt honh giao im ca (d) v (P) l: 2 24 2 5 4 2 5 0 1x m x m x m x m

2 2 24 4 2 5 4 4 2 5 4 2 2m m m m m m m

(d) ct (P) ti hai im phn bit khi Pt (1) c hai nghim phn bit khi 0

2 0

2 0 2 0 22 2 0

2 0 22 0

2 0

m

m m mm m

m mm

m

Vy: vi m > 2 hoc m < -2 th (d) ct (P) ti hai im phn bit.

b) Tm cc gi tr ca m sao cho 3 31 2 0x x .

Vi m > 2 hoc m < -2. Th Pt: 2 4 2 5 0 1x m x m c hai nghim phn bit x1, x2

Theo Viet ta c: 1 2

1 2

4

2 5

x x m

x x m

Ta c 2 23 3

1 2 1 2 1 2 1 23 4 4 3 2 5x x x x x x x x m m m

2

4 1m m .

: 3 31 2 0x x 2

4 1 0 4m m m (tha mn iu kin) hoc 1m (khng tha mn iu

kin)

Vy : 4m l gi tr cn tm.

THY HONG XUN VNH 11


Bi 3: (1,5 im )

Ta c : 18x y

K: 0 0;x y

Pt vit: 3 2x y (1) ( Vi K: 0 0;x y 0 0;x y m 3 2x y => 3 2x v

3 2y )

Pt vit: 3 2 0x y 2 2 18

3 2 6 2 18 26

y xx y y y x y Q

22 2 02 22

a N vi y Z va ay a Q y a Q

a

2a m m N Vy: 2 22 2 2 2y m y m y m . Tng t: 2x n

Pt (1) vit: 2 2 3 2 3 ,n m n m voi m n N

0

3

n

m

hoc

1

2

n

m

hoc

2

1

n

m

hoc

3

0

n

m

0

18

x

y

hoc

2

8

x

y

hoc

8

2

x

y

hoc

18

0

x

y

Vy Pt cho c 4 nghim0

18

x

y

;

2

8

x

y

;

8

2

x

y

;

18

0

x

y

Bi 4: ( 3,5 im )Cho ng trn (O) v mt im P ngoi ng trn. K hai tip tuyn PA, PB vi ng

trn (O) (A ,B l hai tip im). PO ct ng trn ti hai im K v I ( K nm gia P v O) v ct AB ti H. Gi D

l im i xng ca B qua O, C l giao im ca PD v ng trn (O).

a.Chng minh t gic BHCP ni tip. b.Chng minh .AC CH

c.ng trn ngoi tip tam gic ACH ct IC ti M. Tia AM ct IB ti Q. Chng minh M l trung im ca AQ.

Bi 4: ( 3,5 im )

a) Chng minh t gic BHCP ni tip

Xt ABP c: PA = PB

v APO OPB (tnh git hai tip tuyn ct nhau)

=> ABP cn ti P c PO l phn gic

=> PO cng l ng cao, trung tuyn ABP .

Xt t gicBHCP ta c 090BHP (V PO )AB

1

1

Q

M

CD

H

I

K

P

O

A

B

THY HONG XUN VNH 12


090BCP

(V k b 090BCD (ni tip na ng trn (O))

BHP BCP

=> T gicBHCP ni tip (Qi tch cung cha gc)

b) Chng minh .AC CH

Xt ACH ta c

1HAC B (chn cung BKC ca ng trn (O))

M 1 1

B H ( do BHCP ni tip)

=> 1

HAC H

M 01

90H AHC ( V: PO AB)

=> 090HAC AHC

=> AHC vung ti C

Hay .AC CH

c) Chng minh M l trung im ca AQ.

Xt t gic ACHM ta c M nm trn ng trn ngoi tip ACH )

=> t gic ACHM ni tip

=> CMH HAC (chn cung HC )

M HAC BIC (chn cung BC ca ng trn (O))

=> CMH BIC

=> MH//BI (v cp gc ng v bng nhau)

Xt ABQ c AH = BH ( do PH l trung tuyn APB (C/m trn))

V: MH//BI

=> MH l trung bnh ABQ

=> M l trung im ca AQ

Bi 5: (1,0 im)

Ta c: 2 1 2 1 2 1

2 1 3 31 1 1

x xy

x x x x x x

THY HONG XUN VNH 13


V 0< x 2

01

x

x

v

10

x

x

Ta c: 2 1 2 1

2 2 21 1

.x x x x

x x x x

(Bt ng thc C si)

Du = xy ra khi: 2 1

1

x x

x x

2

12 1 0 1 2x x x (tha mn iu kin)

2

1 2x (khng tha mn iu kin; loi)

=> 2 2 3y Du = xy ra khi 1

1 2x

Vy 2 2 3nhonhat

y khi 1

1 2x

4 S GIO DC V O TO K THI TUYN SINH LP 10 THPT

TP. NNG Nm hoc: 2014 2015

CHNH THC MN: TON


Bi 1: (1,5 im)

1) Tnh gi tr ca biu thc 9 4A

Rt gn biu thc 2 2 2

22 2

x xP

xx x

, vi x > 0, 2x

Bi 2: (1,0 im)

Gii h phng trnh 3 4 5

6 7 8

x y

x y

Bi 3: (2,0 im) Cho hm s y = x2 c th (P) v hm s y = 4x + m c th (dm)

1)V th (P) 2)Tm tt c cc gi tr ca m sao cho (dm) v (P) ct nhau ti hai im phn bit, trong tung ca mt

trong hai giao im bng 1.

Bi 4: (2,0 im) Cho phng trnh x2 + 2(m 2)x m2 = 0, vi m l tham s.

1)Gii phng trnh khi m = 0. 2)Trong trng hp phng trnh c hai nghim phn bit x1 v x2 vi x1 < x2, tm tt c cc gi tr ca m

sao cho 1 2 6x x

Bi 5: (3,5 im) Cho tam gic ABC vung ti A c ng cao AH (H thuc BC). V ng trn (C) c tm C, bn knh CA. ng thng AH ct ng trn (C) ti im th hai l D. 1)Chng minh BD l tip tuyn ca ng trn (C).

2)Trn cung nh AD ca ng trn (C) ly im E sao cho HE song song vi AB. ng thng BE ct ng trn (C) ti im th hai l F. Gi K l trung im ca EF. Chng minh rng:

a) BA2 = BE.BF v BHE BFC b) Ba ng thng AF, ED v HK song song vi nhau tng i mt.

THY HONG XUN VNH 14


BI GII

Bi 1:

1)A = 3 2 = 1 2)Vi iu kin cho th

2 22 2

12 22 2 2 2

xx xP

x xx x x x

Bi 2:

3 4 5 6 8 10 2 1

6 7 8 6 7 8 6 7 8 2

x y x y y x

x y x y x y y

Bi 3:

1)

2) Phng trnh honh giao im ca y = x2 v ng thng y = 4x + m l : x2 = 4x + m x2 4x m = 0 (1)

(1) c 4 m

(dm) v (P) ct nhau ti hai im phn bit th 0 4 0 4m m

y = 4x + m = 1 => x = 1

4

m

Yu cu ca bi ton tng ng vi

4 4 4

1 7 72 4 4 4

4 4 4

m m m

haym m mm m m

4

7

74

4

m

m

mm

(loi) hay

4

7

4 4 7

m

m

m m

2 24 4 4

5 hay 35 hay 3 16 4 14 49 2 15 0

m m mm m

m mm m m m m

THY HONG XUN VNH 15


Bi 4:

1)Khi m = 0, phng trnh thnh : x2 4x = 0 x = 0 hay x 4 = 0 x = 0 hay x = 4

2) 2 22 2 22 2 4 4 2 2 1 2 2 1 2 0m m m m m m m m Vy phng trnh lun c hai nghim phn bit vi mi m.

Ta c 21 2 1 22 2 , 0 S x x m P x x m

Ta c 22 2

1 2 1 1 2 2 1 2 1 2 1 26 2 36 2 2 36 x x x x x x x x x x x x

2 2

4 2 36 2 9 m m 1hay 5 m m

Khi m = -1 ta c 1 2 1 2x 3 10,x 3 10 x x 6 (loi)

Khi m = 5 ta c 1 2 1 2x 3 34,x 3 34 x x 6 (tha)

Vy m = 5 tha yu cu bi ton.

Bi 5:

1)Ta c0BAC 90 nn BA l tip tuyn vi (C).

BC vung gc vi AD nn

H l trung im AD. Suy ra 0BDC BAC 90

nn BD cng l tip tuyn vi (C)

2)

a)

Trong tam gic vung ABC

ta c 2AB BH.BC (1)

Xt hai tam gic ng dng ABE v FBA

v c gc B chung

v BAE BFA (cng chn cung AE)

suy ra 2AB BE AB BE.FB

FB BA (2)

T (1) v (2) ta c BH.BC = BE.FB

T BE.BF= BH.BCBE BH

BC BF

2 tam gic BEH v BCF ng dng v c gc B chung v BE BH

BC BF

BHE BFC

THY HONG XUN VNH 16


b) do kt qu trn ta c BFA BAE

HAC EHB BFC , do AB //EH. suy ra DAF DAC FAC DFC CFA BFA

DAF BAE , 2 gc ny chn cc cung AE,DF nn hai cung ny bng nhau

Gi giao im ca AF v EH l N. Ta c 2 tam gic HED v HNA bng nhau

(v gc H i nh, HD = HA, EDH HDN (do AD // AF)

Suy ra HE = HN, nn H l trung im ca EN. Suy ra HK l ng trung bnh ca tam gic EAF.

Vy HK // AF.

Vy ED // HK // AF

.--------------------------------------------------------------------------

5

S GIO DC V O TO NAM NH

THI TUYN SINH LP 10 TRNG THPT CHUYN Nm hc 2014 2015

Mn: TON (chung)

Thi gian lm bi: 120 pht. ( thi gm 01 trang)

Bi 1: (1,5 im):

1) Tm iu kin xc nh ca biu thc 2x

2) Tm bn knh ng trn ngoi tip tam gic vung c di cnh huyn l 10cm.

3) Cho biu thc 2 4 2P x x . Tnh gi tr ca P khi 2x .

4) Tm ta ca im thuc parbol y = 2x2 bit im c honh x = 1.

Bi 2: (1,5 im):

Cho biu thc 2 1 1 2

1 1 1

a a aQ

a a a a a a

vi 0; 1a a .

1) Rt gn biu thc Q.

2) Chng minh rng khi a > 1 th gi tr biu thc Q nh hn 1.

Bi 3: (2,5 im):

1) Cho phng trnh 2 2 2 0 ( )x x m ( m l tham s).

a) Tm m phng trnh (*) c nghim.

CHNH THC

THY HONG XUN VNH 17


b) Gi s 1 2;x x l hai nghim ca phng trnh (*). Tm gi tr nh nht ca biu thc:

2 2 2 21 2 1 23 4A x x x x

2) Gii h phng trnh:

3

3 3

2 1 5 5

1.

x y x

x y

.

Bi 4: (3,0 im): Cho hai ng trn 1 1;O R v 2 2;O R vi 1 2R R tip xc trong vi nhau ti A. ng

thng 1 2O O ct 1 1;O R v 2 2;O R ln lt ti B v C khc A. ng thng i qua trung im D ca BC vung

gc vi BC ct 1 1;O R ti P v Q.

1) Chng minh C l trc tm tam gic APQ.

2) Chng minh 2 2 2

1 2 .DP R R

3) Gi s 1 2 3 4; ; ;D D D D ln lt l hnh chiu vung gc ca D xung cc ng thng ; ; ;BP PA AQ QB .

Chng minh 1 2 3 41

2DD DD DD DD BP PA AQ QB

Bi 5: (1,5 im):

1) Gii phng trnh 2 1 2 1 1.x x x

2) Xt cc s thc x, y, z tha mn 2 2 22 3 36y yz z x . Tm gi tr ln nht v gi tr nh nht ca biu thc .A x y z

Ht

HD mt s cu:

Bi 3:

2)

3

3 3

2 1 5 5 1

1 2

x y x

x y

tr tng v tng ng ca (1) v (2) ta c

3 3 2 2 2 25 0 5 0 5 0(3)x y

x y x y x y x xy yx xy y

PT (3)

2

21 3 5 02 4

x y y

v nghim

THY HONG XUN VNH 18


Vi 3

3 33

1 42 2 1

2 2x y x y x y .

Vy hpt c nghim duy nht 3 34 4

; ;2 2

x y

Bi 4:

1) PBQC l hnh thoi => QC // BP

CM // BP (cng vung gc vi PA)

=> Q, C, M thng hng

Tam gic APQ c 2 ng cao AD v QM

ct nhau ti C

=> C l trc tm tam gic APQ

2) c/minh DM l tip tuyn ti M ca (O2)

Cminh c PD2 = DB.DA = DC.DA = DM2

= O2D2 O2M2 = O2D2 R22

Ta i cminh O2D = R1

Ta c 12 2 12

2 2 2 2

AC BC AB RO D O A CD R

Vy ta c pcm.

c) 1 2 3 41

2DD DD DD DD BP PA AQ QB

D dng cminh c 1 4 2 3; ; ;DD DD DD DD BP QB PA AQ

Nn 1 2 3 4 1 21

22

DD DD DD DD BP PA AQ QB DD DD PB PA

Ap dng BT C-si ta c 2 2 2 2 2 22 . 2 . ( )DB DP DB DP BP DB DP Pi ta go DB DP BP

1

2 .2DD

DB DPBP

BP (du = xy ra khi DP = DB) (1)

D4

D3

D2

D1

O2O1

M

Q

P

D

CB A

THY HONG XUN VNH 19


Cminh tng t ta c2

2 .2DD

DA DPAP

AP (du = xy ra khi DP = DA) (2)

T (1) v (2) => 1 22 DD DD PB PA (du = xy ra khi DP = DA =DB)

Bi 5:

1) KX 21 x

2 1 2 1 1.x x x

1 3 12 1

2 1 2 1 2 1

2 1 3 2 3 2 2 1 1 3 2

2 2 1 13 2 2 3 2 (*)

2 2 1 1 2 2 1 1

x xx x x x

x x x x x x

x xx x x

x x x x

Xt PT (*) ta c:

+) x = 2 tha mn

+) 1 x < 2 V tri m v phi dng V l !

+) x > 2 khng thuc KX

Vy x = 2 l nghim PT cho

2) Ta c:

2 2 2 2

2 2 2 2 2 2 2

2 2 2

2 2 2

2( ) 3 ( 2 ) ( 2 )

36 ( ) ( ) 36

x y z x y z xy yz xz

y z yz x x xy y x xz z

x y z x y x z

Nn 66 zyx

=> Max(x+y+z) = 6 khi x = y = z = 2

Min(x+y+z) = 6 khi x = y = z = 2

----------------------------------------------------------------------------------------------------------------------------------

6

S GIO DC V O TO

HI DNG

K THI TUYN SINH LP 10

THPT CHUYN NGUYN TRI

NM HC 2014 2015

Mn thi: Ton ( khng chuyn )

THY HONG XUN VNH 20


CHNH

THC


thi gm: 01 trang

Cu I ( 2,0 im)

1) Gii phng trnh: 43 1x x

2) Rt gn biu thc: 10 2 3 1

( 0; 1)3 4 4 1

x x xA x x

x x x x

Cu II ( 2,0 im)

Cho Parabol (P): 2y x v ng thng (d): ( 1) 4y m x m (tham s m)

1) Vi m = 2, tm ta giao im ca (P) v (d). 2) Tm m (d) ct (P) ti hai im nm v hai pha ca trc tung.

Cu III ( 2,0 im)

1) Cho h phng trnh: 3 2

3 2 11

x y m

x y m

( tham s m)

Tm m h cho c nghim (x; y) tha mn x2 y2 t gi tr ln nht.

2) Mt t d nh i t A n B di 80 km vi vn tc d nh. Thc t trn na

qung ng u t i vi vn tc nh hn vn tc d nh l 6 km/h. Trong na qung

ng cn li t i vi vn tc nhanh hn vn tc d nh l 12 km/h. Bit rng t n

B ng thi gian nh. Tm vn tc d nh ca t.

Cu IV ( 3,0 im)

Cho tam gic ABC nhn, cc ng cao AM, BN, CP ca tam gic ABC ct nhau ti

H. Dng hnh bnh hnh BHCD.

1) Chng minh: Cc t gic APHN, ABDC l cc t gic ni tip. 2) Gi E l giao im ca AD v BN. Chng minh: AB.AH = AE.AC

3) Gi s cc im B v C c nh, A thay i sao cho tam gic ABC nhn v BAC

khng i. Chng minh rng ng trn ngoi tip t gic APHN c din tch khng i.

THY HONG XUN VNH 21


Cu V ( 1,0 im)

Cho x; y l hai s dng thay i. Tm gi tr nh nht ca biu thc:

2 2

2 2

x y x yS

x y xy

-----------------------------Ht------------------------------

H v tn th sinh :S bo danh :...

Ch k ca gim th 1 :Ch k ca gim th 2 :...

S GIO DC V O TO

HI DNG

HNG DN CHM

THI TUYN SINH LP 10 THPT

CHUYN

NGUYN TRI NM HC 2014 2015

Mn thi: Ton ( khng chuyn )

I) HNG DN CHUNG

- Th sinh lm bi theo cch khc nhng ng vn cho im ti a.

- Sau khi cng im ton bi, im l n 0,25 im.

II) P N V BIU IM CHM

Cu Ni dung im

I 1 Gii phng trnh: 43 1x x 1,00

2

1 0 (1)43 1

43 1 (2)

xx x

x x

0,25

(1) 1x 0,25

(2)

2 42 0x x 7

6

x

x

0,25

THY HONG XUN VNH 22


Kt hp nghim ta c 7x (tha mn), 6x ( loi)

Vy tp nghim phng trnh cho l 7S

0,25

I

2

Rt gn biu thc:

10 2 3 1 ( 0; 1)

3 4 4 1

x x xA x x

x x x x

1,00

10 2 3 1

4 14 1

x x xA

x xx x

0,25

10 2 3 1 1 4

4 1

x x x x x

x x

0,25

10 2 5 3 5 4 3 10 7 =

4 1 4 1

x x x x x x x

x x x x

0,25

1 7 3 7 3 = =

44 1

x x x

xx x

( v 0; 1x x )

0,25

II

Cho Parabol 2: P y x v ng thng

: ( 1) 4d y m x m

(tham s m)

2,00

1 Vi m = 2, tm ta giao im ca (P) v (d). 1,00

m = 2 ta c phng trnh ng thng (d) l: y = x + 6 0,25

Honh giao im ca (P) v (d) l nghim ca phng trnh

2 6x x

0,25

2

2 6 0

3

xx x

x

0,25

* 2 4x y

* 3 9x y

0,25

THY HONG XUN VNH 23


Vy m = 2 th (P) v (d) ct nhau ti hai im 2;4A v

3;9B

II 2 Tm m (d) ct (P) ti hai im nm v hai pha ca trc tung. 1,00

Honh giao im ca (P) v (d) l nghim ca phng trnh

2 1 4x m x m

2 1 4 0x m x m (*)

0.25

(d) ct (P) ti hai im nm v hai pha ca trc tung khi v

ch khi phng trnh (*) c hai nghim tri du

0,25

1. 4 < 0 m 0,25

m > 4 0,25

III

1 Cho h phng trnh:

3 2

3 2 11

x y m

x y m

( tham s m)

1,00

Gii h phng trnh ta c

3

2 1

x m

y m

0,25

2 22 2 23 2 1 = 3 10 8x y m m m m

249 5

= 33 3

m

0,25

Do

25

03

m

vi mi m; du = xy ra khi5

3

m

0,25

2 2 49 3

x y , du = xy ra khi5

3

m

hay 2 2 x y ln nht bng 49

3 khi

5

3m

0,25

III 2 Gi vn tc d nh ca t l x (km/h) (x >6 )

Khi thi gian t d nh i ht qung ng AB l 80

( )hx

0,25

THY HONG XUN VNH 24


Thi gian thc t t i na qung ng u l

40( )

6h

x

Thi gian thc t t i na qung ng cn li l 40

( )12

hx

0,25

Theo bi ra ta c phng trnh:

40 40 80

6 12x x x

0,25

Gii phng trnh ta c 24x ( tha mn)

Vy vn tc d nh ca t l 24 (km/h)

0,25

IV

1

T gi thit ta c 090APH v 090ANH

0,25

t gic APHN ni tip ng trn (ng knh AH) 0,25

Ta c : BD// CH ( BDCH l hnh bnh hnh) v CH AB

BD AB 090ABD

Tng t c 090ACD

0,25

t gic ABDC ni tip ng trn ( ng knh AD ) 0,25

IV 2 Xt 2 tam gic ABE v ACH c :

ABE ACH ( cng ph vi BAC ) (1)

0,25

BAE ph vi BDA ; BDA BCA (gc nt cng chn AB )

CAH ph vi BCA

BAE CAH (2)

0,25

T (1) v (2) suy ra 2 tam gic ABE, ACH ng dng 0,25

I

O

E

M

D

N

P

CB

A

H

THY HONG XUN VNH 25


. .

AB ACAB AH AC AE

AE AH

0,25

IV 3 Gi I l trung im BC I c nh (Do B v C c nh) 0,25

Gi O l trung im AD O c nh ( Do BAC khng i,

B v C c nh, O l tm ng trn ngoi tip tam gic ABC )

di OI khng i

0,25

ABDC l hnh bnh hnh I l trung im HD

1

2OI AH ( OI l ng trung bnh tam gic ADH)

di AH khng i

0,25

V AH l ng knh ng trn ngoi tip t gic APHN,

di AH khng i di bn knh ng trn ngoi tip t

gic APHN khng i ng trn ngoi tip t gic APHN

c din tch khng i.

0,25

V Ta c:

2 2

2 2

x y x yS

x y xy

2 2

2 2

2 1+ 2

xy x y

x y xy

0,25

2 2 2 2

2 2

2 3+

2 2

xy x y x y

x y xy xy

0,25

Do x; y l cc s dng suy ra

2 2 2 2

2 2 2 2

2 2 2 . 2

2 2

xy x y xy x y

x y xy x y xy

; =

2 2

2 22 2 2 2 2 2

2 2

24 0

2

x y xyx y x y x y

xy x y

2 2 ( ; 0)x y x y x y

0,25

THY HONG XUN VNH 26


2 22 2 2 1

2

x yx y xy

xy

; = x y

Cng cc bt ta c 6S

6 S x y .Vy Min S = 6 khi v ch khi x = y

0,25

7

S GIO DC V O TO TY NINH K THI TUYN SINH VO LP 10 NM HC 2014 2015

Ngy thi : 21 thng 6 nm 2014

Mn thi : TON (Khng chuyn)

Thi gian : 120 pht (Khng k thi gian giao )

-------------------------------------------------------------------------------------

CHNH THC

( thi c 01 trang, th sinh khng phi chp vo giy thi)

Cu 1 : (1im) Thc hin cc php tnh

a) A 2 5 2 5 b) B = 2 50 3 2

Cu 2 : (1 im) Gii phng trnh: 22 15 0x x .

Cu 3 : (1 im) Gii h phng trnh:

23

12 4

yx

yx

.

Cu 4 : (1 im) Tm a v b ng thng d : a 2 by x c h s gc bng 4 v i qua

im M 1; .

THY HONG XUN VNH 27


Cu 5 : (1 im) V th ca hm s 22y x .

Cu 6 : (1 im) Lp 9A d nh trng 420 cy xanh. n ngy thc hin c 7 bn khng tham

gia do c triu tp hc bi dng i tuyn hc sinh gii ca nh trng nn mi bn cn li

phi trng thm 3 cy mi m bo k hoch t ra. Hi lp 9A c bao nhiu hc sinh.

Cu 7 : (1 im) Chng minh rng phng trnh 2 2 m +1 m 4 0x x lun c hai nghim

phn bit 1x , 2x v biu thc 1 2 2 1M 1 1x x x x khng ph thuc vo m.

Cu 8 : (2 im) Cho tam gic ABC vung ti A c ng cao AH (H thuc BC), bit 0ACB 60

, CH = a . Tnh AB v AC theo a.

Cu 9 : (1 im) Cho ng trn tm O ng knh AB c nh, CD l ng knh thay i ca

ng trn (O) (khc AB). Tip tuyn ti B ca (O) ct AC v AD ln lt ti N v M. Chng

minh t gic CDMN ni tip.

Cu 10 : (1 im) Cho t gic ABCD ni tp ng trn tm O, bn knh bng a. Bit AC vung

gc vi BD. Tnh 2 2AB CD theo a.

--- HT ---

Gim th khng gii thch g thm.

H v tn th sinh : ................................................ S bo danh : ......................................

Ch ky ca gim th 1: ......................................... Ch ky ca gim th 2 : ........................

BI GII

Cu 1 : (1im) Thc hin cc php tnh

a) 2

2A 2 5 2 5 2 5 4 5 1 .

b) B = 2 50 3 2 100 3.2 10 6 4 .

Cu 2 : (1 im) Gii phng trnh: 22 15 0x x .

21 4.2. 15 121 0 , 11 .

1

1 11 10 5

4 4 2x

; 2

1 11 123

4 4x

.

Vy 5

S = ; 32

.

THY HONG XUN VNH 28


Cu 3 : (1 im) iu kin 0x .

23

12 4

yx

yx

42 6

12 4

yx

yx

510

23

x

yx

5

10

23

x

yx

1

2

4 3

x

y

1

2

1

x

y

(nhn).

Vy h phng trnh c nghim duy nht 1

; ; 12

x y

.

Cu 4 : (1 im) Tm a v b d : a 2 by x c h s gc bng 4 v qua M 1; .

ng thng d c h s gc bng 4 a 2 4 a 6 .

Mt khc (d) i qua im M 1; nn thay a 6 , 1x ; 3y vo a 2 by x .

Khi ta c : 3 6 2 .1 b 3 4 b b 7 .

Vy a 6 v b 7 l cc gi tr cn tm v khi d : 6 7y x .

Cu 5 : (1 im) V th ca hm s 22y x .

BGT

x 2 1 0 1 2 22y x 8 2 0 2 8

Cu 6 : (1 im)

Gi s hc sinh lp 9A l x , 7x x .

Theo k hoch, mi em phi trng 420

x (cy).

THY HONG XUN VNH 29


Trn thc t. s hc sinh cn li l : 7x .

Trn thc t, mi em phi trng 420

7x (cy).

Do lng cy mi em trng trn thc t hn 3 cy so vi k hoch nn ta c phng trnh :

420 420

3 77

xx x

420 420 7 3 7x x x x

23 21 2940 0x x

2 7 980 0x x (chia 3)

27 4.1. 980 3969 0 , 3969 63 .

1

7 6335

2x

(nhn) ; 2

7 6328

2x

(loi).

Vy lp 9A c 35 hc sinh.

Cu 7 : (1 im) Phng trnh 2 2 m +1 m 4 0x x .

Phng trnh c 2 2 2' m 1 1. m 4 m 2m 1 m 4 m m 5 .

2 2

2 1 1 1 19' m m 5 m 5 m 0, m2 4 2 4

.

Vy phng trnh lun c hai nghim phn bit vi mi m.

Khi , theo Vi-t 1 2 2m 2x x ; 1 2. m 4x x .

1 2 2 1 1 1 2 2 1 2 1 2 1 2M 1 1 2x x x x x x x x x x x x x x .

1 2 1 2M 2 2m 2 2 m 4 2m 2 2m 8 10x x x x (khng ph thuc vo m).

Cu 8 :

GT ABC , 0A 90 , AH BC , 0ACB 60 , CH = a

KL Tnh AB v AC theo a?

ACH c CH

cosCAC

nn 0

CH a aAC 2a

1cosC cos60

2

.

THY HONG XUN VNH 30


ABC c 0AB = AC.tanC = 2a.tan 60 2a. 3 2 3a .

Vy AB = 2 3a , AC 2a .

Cu 9 : (1 im)

GT (O) ng knh AB c nh, ng knh CD thay i, MN l tip tuyn ti B ca (O).

KL T gic CDMN ni tip

Chng minh t gic CDMN ni tip

Ta c : 1

ADC AC2

s .

1 1 1N ADB BC ACB BC AC2 2 2

s s s s s .

ADC N (cng bng 1

AC2s ).

T gic CDMN ni tip c (gc ngoi bng gc i trong).

Cu 10 : (1 im)

GT ABCD ni tip O; a , AC BD KL Tnh

2 2AB CD theo a.

Tnh 2 2AB CD theo a.

V ng knh CE ca ng trn (O).

Ta c : 0EAC 90 , 0EDC 90 (gc ni tip chn ng knh EC).

THY HONG XUN VNH 31


AC AEAE BD

AC BD ( )gt

ABDE l hnh thang cn (hnh thang ni tip (O))

AB= DE (cnh bn hnh thang cn).

22 2 2 2 2 2AB + CD = DE + DC = EC 2a 4a (do EDC vung ti D).

Vy 2 2 2AB CD 4a .

8

S GIO DC V O TO

QUNG NGI

K THI TUYN SINH VO LP 10 THPT

NM HC 2014-2015

MN : TON (khng chuyn)

Ngy thi: 19/6/2014

Thi gian lm bi: 120 pht (khng k thi gian giao )

Bi 1: (1,5 im)

a/ Tnh: 43252

b/ Xc nh a v b th hm s y = ax + b i qua im A(1; 2) v im B(3;

4)

c/ Rt gn biu thc A = 2x

4x:

2x

2

2x

x

vi x 0 v x 4

Bi 2: (2,0 im)

1/ Gii phng trnh x4 + 5x2 36 = 0

2/ Cho phng trnh x2 (3m + 1)x + 2m2 + m 1 = 0 (1) vi m l tham s.

a/ Chng minh phng trnh (1) lun c hai nghim phn bit vi mi gi tr ca

m.

b/ Gi x1, x2 l cc nghim ca phng trnh (1). Tm m biu thc

B = x12 + x22 3x1x2 t gi tr ln nht.

Bi 3: (2,0 im)

CHNH THC

THY HONG XUN VNH 32


chun b cho mt chuyn i nh bt c Hong Sa, hai ng dn o L

Sn cn chuyn mt s lng thc, thc phm ln tu. Nu ngi th nht

chuyn xong mt na s lng thc, thc phm; sau ngi th hai chuyn

ht s cn li ln tu th thi gian ngi th hai hon thnh lu hn ngi th

nht l 3 gi. Nu c hai cng lm chung th thi gian chuyn ht s lng thc,

thc phm ln tu l 7

20 gi. Hi nu lm ring mt mnh th mi ngi chuyn

ht s lng thc, thc phm ln tu trong thi gian bao lu?

Bi 4: (3,5 im)

Cho na ng trn tm O, ng knh AB = 2R. Gi M l im chnh gia

ca cung AB; P l im thuc cung MB (P khc M v P khc B). ng thng AP

ct ng thng OM ti C; ng thng OM ct ng thng BP ti D. Tip tuyn

ca na ng trn P ct ct CD ti I.

a/ Chng minh OADP l t gic ni tip ng trn.

b/ Chng minh OB.AC = OC.BD.

c/ Tm v tr ca im P trn cung MB tam gic PIC l tam gic u. Khi

hy tnh din tch ca tam gic PIC theo R.

Bi 5: (1,0 im)

Cho biu thc A = (4x5 + 4x4 5x3 + 5x 2)2014 + 2015. Tnh gi tr ca biu

thc A khi x = 12

12

2

1

.

----------------------------------- HT -------------------------------

Gim th coi thi khng gii thch g thm

GI BI GII TON VO 10 KHNG CHUYN L KHIT QUNG NGI.

THY HONG XUN VNH 33


Bi 1: a/ Tnh: 43252 = 10 + 6 = 16

b/ th hm s y = ax + b i qua A(1; 2) nn a + b = 2, v B(3; 4) nn 3a b = 4.

Suy ra a = 3, b = 5. Vy (d): y = 3x + 5

c/ Vi x 0 v x 4 ta c:A = 2x

4x:

2x

2

2x

x

= ..=

4x

2x

2x

1

Bi 2:

1/ Gii phng trnh x4 + 5x2 36 = 0

t t = x2 ( t 0) ta c phng trnh t2 + 5t 36 = 0. t = 25 4.1.(36) = 169

t1 = 4 (tmk); t2 = 9 (loi). Vi t = 4 x2 = 4 x = 2

2/ a/ Vi m l tham s, phng trnh x2 (3m + 1)x + 2m2 + m 1 = 0 (1)

C = [(3m + 1)]2 4.1.( 2m2 + m 1) = m2 + 2m + 5 = (m + 1)2 + 4 > 0 m

Vy phng trnh (1) lun c hai nghim phn bit vi mi gi tr ca m.

b/ Gi x1, x2 l cc nghim ca phng trnh (1). Ta c x1 + x2 = 3m + 1; x1x2 = 2m2 + m 1

B = x12 + x22 3x1x2 = (x1 + x2)2 5x1x2 = (3m + 1)2 5(2m2 + m 1) = (m2 m 6)

B = (m 2

1)2 +

2

13

2

13. Du = xy ra m

2

1 = 0 m =

2

1.

Vy Bmin = 2

13 khi m =

2

1

Bi 3: Gi x (gi) l thi gian ngi th I mt mnh lm xong c cng vic.

v y (gi) l thi gian ngi th II mt mnh lm xong c cng vic. (Vi x, y > 7

20)

Ta c h phng trnh:

32

x

2

y

20

7

y

1

x

1

)2(6xy

)1(20

7

y

1

x

1

T (1) v (2) ta c phng trnh: 20

7

6x

1

x

1

. Gii phng trnh c x1 = 4, x2 =

7

30

Chn x = 4.

Vy thi gian mt mnh lm xong c cng vic ca ngi th I l 4 gi, ca ngi th II l 10 gi.

Bi 4:

a/ C/minh AOD = APD = 900

THY HONG XUN VNH 34


O v P cng nhn on AD di mt gc 900

OADP t gic ni tip ng trn ng knh AD

b/ C/ minh AOC DOB (g.g) DB

AC

OB

OC

OB.AC = OC.BD (pcm)

c/ Ta c IPC = PBA (cng chn cung AP ca (O))

v c ICP = PBA (cng b vi OCP)

Suy ra IPC = ICP IPC cn ti I.

IPC l tam gic u th IPC = 600 PBA = 600

OP = PB = OB = R s o cung PB bng 600

C/minh DIP cn ti I ID = IP = IC = CD:2

Do SPIC = 2

1SDPC =

2

1.2

1.CP.PD =

4

1.

3

3R.R =

12

3R 2 (vdt)

Bi 5:

Ta c: x =12

12

2

1

=

12.12

12

2

12

=

2

12

x2 = 4

223; x3 = x.x2 =

8

725 ; x4 (x2)2 =

16

21217 ; x5 = x.x4 =

32

41229

Do : 4x5 + 4x4 5x3 + 5x 2 = 18

1620220352252243441229

Vy A = (4x5 + 4x4 5x3 + 5x 2)2014 + 2015 = (1)2014 + 2015 = 1 + 2015 = 2016

---------------------------------------------------------------

9

P

D

I

M

C

OBA

S GIAO DC V O TO

PHU TH

CHINH THC

K THI TUYN SINH

VO LP 10 TRUNG HC PH THNG

NM HC 2014-2015

Mn toan

Thi gian lm bi: 120 pht, khng k thi gian giao

thi co 01 trang

-------------------------------------------

THY HONG XUN VNH 35


Cu1 (1,5im)

a) Trong cc phng trnh di y, nhng phng trnh no l phng trnh bc 2:

012)1(

012

043

;023

2

2

2

mxxm

x

x

xx

( x l n s m l tham s m khc 1)

b)Gii phng trnh : 642 x

Cu2 (2,0 im)

a) Gii h phng trnh

3

53

yx

yx

b) Rt gn biu thc ba

ba

ab

abbaB

,vi a,b l s dng.

Cu3 (2,0 im)

Cho phng trnh bc 2: 0)12( 22 mxmx (1)

a) Gii phng trnh vi m = 1 b) Vi gi tr no ca m phng trnh (1) c nghim kp.Tm nghim kp

Cu 4( 3,0 im)

Cho (O;R) Dy BC

THY HONG XUN VNH 36


Ghi ch : Cn b coi thi khng gii thch g thm

Hng n

Cu1 (1,5im)

a) Cc phng trnh 012)1(;043;023 222 mxxmxxx

b)Gii phng trnh : 282642 xxx

Cu2 (2,0 im)

a) Gii h phng trnh

2

1

3

22

3

53

y

x

xy

x

yx

yx

b) Rt gn biu thc

ababaB

ba

baba

ab

baab

ba

ba

ab

abbaB

2

)(

,

vi a,b l s dng.

Cu3 (2,0 im)

Cho phng trnh bc 2: 0)12( 22 mxmx (1)

a)Gii phng trnh vi m = 1:Thay m=1 ta c PT : 0132 xx

543 2 PT C 2 nghim 2

53;

2

5321

xx

b) Vi gi tr no ca m phng trnh (1) c nghim kp.Tm nghim kp

4

1

4

10

144144412 2222

xm

mmmmmm

THY HONG XUN VNH 37


Cu 4 (3,0 im)

Hng dn

a) Dng nh l o v I l trung im AH b) Gi M l trung im BC Chng minh ME l tip tuyn (I) c) K ng knh AK ta c BHCK l hnh bnh hnh ( theo nh ngha) nn H,M, K

thng hng Xt tam gic AHK c OM l ng trung bnh suy ra AH=2.OM khng i dng trn ngoi tip tam gic AEF nhn AH l ng knh c bn knh bng OM khng i

Tam gic AEF ng dng vi tam gic ABC nn

ABCAEF

ABC

AEF SR

OMS

OA

OM

S

S22

ta c

R

OM khng i

)()()( MaxADMaxSMaxS ABCAEF

M OMOAAMAD ( Khng i) MDOMRAD (max) hay A l chnh gia

cung ln BC

Cu 5(1,5 im) Gii phng trnh 032)52(356 23 xxxxx (1)

Hng dn

KX :2

3x

H

K

O

I

MD

F

E

CB

A

A

THY HONG XUN VNH 38


0321

524)2(

0321

2).52()4)(2(

0321

2).52(824

0)132)(52()1)(52(354

032)52(356

2

22

223

23

23

xx

xxx

xx

xxxx

xx

xxxxx

xxxxxxxx

xxxxx

Vi 2

3x 0

321

524:

xx

xxth

Nn

2

2022

x

xx

Thay vo PT (1) 2x tha mn

GV Nguyn Minh Sang THCS Lm Thao-Ph Th

10

SGIO DC V O TO HI PHNG

CHNH THC

KTHI TUYN SINH VO LP 10 THPT NM HC 2014 - 2015

MN : TON

Thi gian lm bi:120pht (khng k thi gian giao )

GV gii : NGUYN HU BIN - THCS TAM HNG - Thy Nguyn - Hi Phng

I. PHN 1. TRC NGHIM (2,0 im)

Cu 1: iu kin xc nh ca biu thc:

THY HONG XUN VNH 39


p n : D

Cu 2: Hm s no sau y khng phi l hm s bc nht ?

p n : B

Cu 3: H phng trnh C nghim l cp s(x ; y) bng

A. ( 2; 4) B. (6; 2) C. (6; 4) D. (4; 2)

p n : D

Cu 4: Nu x1 ; x2 l cc nghim ca phng trnh x2+x-1 =0 th tng x12+x22 bng:

A. ( 2; 4) B. (6; 2) C. (6; 4) D. (4; 2)

p n : B

Cu 5: Tam gic MNP vung ti M c ng cao MH. Bit MH = 2; NH = 1, x l di MP, ta c :

p n : C

THY HONG XUN VNH 40


I. PHN 2. T LUN (8,0 im)

Bi 1: (1,5 im)

1. Rt gn cc biu thc :

THY HONG XUN VNH 41


THY HONG XUN VNH 42


THY HONG XUN VNH 43


THY HONG XUN VNH 44


11

S GIO DC V O TO K THI TUYN SINH VO LP 10 THPT CHUYN

KHNH HO NM HC 2014 2015

.

MN THI: TON (KHNG CHUYN)

Ngy thi: 20/6/2014

(Thi gian : 120 pht khng k thi gian giao )

Bi 1: (2,00 im)

1) Khng dng my tnh cm tay, tnh gi tr biu thc: 1 8 10

2 1 2 5A

2) Rt gn biu thc B = 1

:2 2 4 4

a a a

a a a a a

vi a > 0, a 4.

Bi 2: (2,00 im)

1) Cho h phng trnh: ax y y

x by a

Tm a v b bit h phng trnh cho c nghim (x, y) = (2; 3).

2)Gii phng trnh: 2 2 1 3 5 6 3 8x x x

Bi 3: (2,00 im) Trong mt phng Oxy cho parabol (P):21

2y x

a)V th (P).

b)Trn (P) ly im A c honh xA = -2. Tm ta im M trn trc Ox sao cho MA MB t gi tr ln

nht, bit rng B(1; 1).

Bi 4: (2,00 im) Cho na ng trn (O) ng knh AB = 2R. V ng thng d l tip tuyn ca (O) ti B.

THI CHNH THC

THY HONG XUN VNH 45


Trn cung AB ly im M ty (M khc A v B), tia AM ct d ti N. Gi C l trung im ca

AM , tia CO ct d ti D.

a) Chng minh rng: OBNC ni tip.

b) Chng minh rng: NO AD

c) Chng minh rng: CA. CN = CO . CD.

d) Xc nh v tr im M (2AM + AN) t gi tr nh nht.

----- HT ----- Gim th khng gii thch g thm.

HNG DN GII

(L Quc Dng, GV THCS Trn Hng o, Nha Trang, Khnh Ho)

Bi 1: (2,00 im)

1) 1 8 10 2 1 2(2 5)

2 1 2 112 1 2 5 2 5

A

2) B = 1

:2 2 4 4

a a a

a a a a a

vi a > 0, a 4.

= 21 ( 2)

:2 2 4 4 2 2 1

a a a a a a

a a a a a a a a

= 2 2( 2) (1 ) ( 2)

( 2)2 1 2 1

a a a a a aa a

a a a a

Bi 2: (2,00 im)

1) V h phng trnh: ax y y

x by a

c nghim (x, y) = (2; 3) nn ta c hpt:

2 3 2 3 6 3 9 7 7 1

2 3 3 2 3 2 2 3 1

a b a b a b a a

b a a b a b a b b

Vy a = 1, b = 1

2) Gii phng trnh: 2 2 1 3 5 6 3 8x x x

THY HONG XUN VNH 46


2 2

4 2 1 6 5 6 2 3 8

((5 6 5 6 9) ((3 2 3 8 1) 0

( 5 6 3) ( 3 8 1) 0

5 6 3 03

3 8 1 0

x 6) x 8)

x x x

x x

x x

xx

x

Vy pt c nghim x = 3.

Bi 3: (2,00 im)

Trong mt phng Oxy cho parabol (P):21

2y x

a)Lp bng gi tr (HS t lm).

th:

b)V A (P) c honh xA = -2 nn yA = 2. Vy A(-2; 2)

Ly M(xM; 0) bt k thuc Ox,

Ta c: MA MB AB (Do M thay i trn Ox v BT tam gic)

Du = xy ra khi 3 im A, B, M thng hng, khi M l giao im ca ng thng AB v trc Ox.

THY HONG XUN VNH 47


- Lp pt ng thng AB

- Tm giao im ca ng thng AB v Ox, tm M (4; 0).

Bi 4: (2,00 im)

Cho na ng trn (O) ng knh AB = 2R. V ng thng d l tip tuyn ca (O) ti B.

Trn cung AB ly im M ty (M khc A v B), tia AM ct d ti N. Gi C l trung im ca

AM , tia CO ct d ti D.

a) Chng minh rng: OBNC ni tip.

HD: T gic OBNC ni tip c 0180OCN OBN

b) Chng minh rng: NO AD

HD: AND co hai ng cao ct nhau ti O,

suy ra: NO l ng cao th ba hay: NO AD

c) Chng minh rng: CA. CN = CO . CD.

d

D

C

N

A OB

M

THY HONG XUN VNH 48


HD: CAO CDN D

CA CO

C CN CA. CN = CO . CD

d) Xc nh v tr im M (2AM + AN) t gi tr nh nht.

Ta c: 2AM + AN 2 2 .AM AN (BT Cauchy Csi)

Ta chng minh: AM. AN = AB2 = 4R2. (1)

Suy ra: 2AM + AN 2 22.4R = 4R 2.

ng thc xy ra khi: 2AM = AN AM = AN/2 (2)

T (1) v (2) suy ra: AM = R 2 M l im chnh gia cung AB

12

thi tuyn sinh vo lp 10 trng chuyn L Qu n, B Ra - Vng Tu nm hoc 2014-2015. Mn Ton chung

Cu 1.

1) Rt gn biu thc2 7 14

A 28 7 52 2

2) Gii h phng trnh 3x 2y 13

2x 3y 12

3) Gii phng trnh x2 5x + 6 = 0.

Cu 2. Cho parabol (P) : 21

y x2

1) V parabol (P). 2) Chng minh rng: Nu ng thng (D): y=x+m i qua im A(4;8) th (D) v (P) khng c im chung.

Cu 3. 1) Cho phng trnh x2 + mxm1=0 (m l tham s). Tm tt c cc gi tr ca m phng trnh trn

c hai nghim x1,x2 tha mn 2 2

1 2 1 2x x 6x x 8

2) Gii phng trnh: 2 2x 2 x 1 2

Cu 4. Cho ng trn (O), ng knh AB v im M c nh thuc ng trn (M khc A v B). D l im di ng trn on thng AM (D khc A v M). ng

THY HONG XUN VNH 49


thng BD ct (O) ti K (K khc B). Hai ng thng AK v BM ct nhau ti C. 1) Chng minh t gic KCMD ni tip.

2) K MHAB ti H. Chng minh 2 2AM.BM

AK BKHM

3) ng thng CD ct AB ti I. Chng minh IC l phn gic ca gc MIK. 4) Xc nh v tr ca im D trn on AM tch DB.DK t gi tr ln nht.

Cu 5 . Cho hai s dng a,b tha mn a+b+ab3. Chng minh bt ng thc :

1 1 1

a b ab 3a b a b 3 4

Gii:

13

S GIO DC V O TO

NGH AN


NM HC 2014 2015

Mn thi: TON

Thi gian lm bi : 120 pht(khng k thi gian giao )

Cu 1. (2,5 im)

Cho biu thc 1 1

:11 1

xA

xx x

a) Nu iu kin xc nh v rt biu thc A

b) Tm tt c cc gi tr ca x 0A .

CHNH THC

THY HONG XUN VNH 50


Cu 2. (1,5 im)

Mt t v mt xe my hai a im A v B cch nhau 180 km, khi hnh cng mt lc

i ngc chiu nhau v gp nhau sau 2 gi. Bit vn tc ca t ln hn vn tc ca xe

my 10 km/h. Tnh vn tc ca mi xe.

Cu 3 . (2,0 im)

Cho phng trnh 2 4 22( 1) 2 0x m x m m (m l tham s)

a) Gii phng trnh khi m = 1.

b) Chng minh rng phng trnh lun c hai nghim phn bit vi mi m.

Cu 4. (3,0 im)

Cho im A nm bn ngoi ng trn (O). T A k hai tip tuyn AB, AC vi ng trn

(B, C l cc tip im). Gi M l trung im ca AB. ng thng MC ct ng trn

(O) ti N (N khc C).

a) Chng minh ABOC l t gic ni tip

b) Chng minh 2 .MB MN MC

c) Tia AN ct ng trn (O) ti D ( D khc N). Chng minh: MAN ADC

Cu 5. (1,0 im)

Cho ba s thc dng , , zx y tha mn x y z . Chng minh rng:

2 2 2 2 2 21 1 1 27

2x y z

x y z

----- Ht ------

AP S CU D V HNG DN GII MT S CU KH

Cu 1. a). iu kin 0

1

x

x

. Rt gn biu thc c:

1

1A

x

b). 0 1x

Cu 2. Vn tc ca t l 50 km/h v vn tc ca xe my l: 40 km/h

Cu 3. a). Phng trnh c hai nghim 1 22 5; 2 5x x

b). Ta c:

THY HONG XUN VNH 51


2 2

4 4 2 2 21 1 1 1' 2m 2 1 2 2 2 2 2 2 0,2 2 2 2

m m m m m m m m

M

2 1 02

' 01

02

m

m

v nghim

Do ' 0, m . Vy phng trnh lun c hai nghim phn bit vi mi m.

Cu 4.

a). Ta c 90 90 180ABO ACO nn t gic ABOC ni tip.

b). Ta c MBN MCB (g-g) nn 2 .MB MN

MB MN MCMC MB

c). Xt MAN v MCA c gc M chung.

V M l trung im ca AB nn MA MB .

Theo cu b ta c: 2 .MA MN MC MA MC

MN MA

Do : MAN MCA (c-g-c)

T suy ra: MAN MCA NCA . (1)

M NCA NDC (gc to bi tip vi dy cung v gc ni tip) (2)

T (1) v (2) suy ra: MAN NDC hay MAN ADC .

Cu 5. Ta c: 2 2 2 2

2 2 2 2

2 2 2 2 2 2 2 2

1 1 1 1 13

x y x yVT x y z z

x y z z x y y x

N

OA

B

C

D

M

THY HONG XUN VNH 52


p dng bt ng thc C si cho hai s dng ta c: 2 2 2 2

2 2 2 22 . 2

x y x y

y x y x

2 2 2 2 2

2 2 2 2 2 2

15 1 15

16 16 16

x z y z zVT

z x z y x y

Li p dng bt ng thc C si ta c: 2 2 2 2

2 2 2 2

12 .

16 16 2

x z x z

z x z x

2 2 2 2

2 2 2 2

12 .

16 16 2

y z y z

z y z y

V 22 2 2

1 1 2 2 8

( )

2

x y xy x yx y

nn 22 2

2 2 2

15 1 1 15 8 15 15.

16 16 ( ) 2 2

z z z

x y x y x y

(v x y z )

Suy ra : 1 1 15 27

52 2 2 2

VT . ng thc xy ra khi 2

zx y .

Vy 2 2 2 2 2 21 1 1 27

2x y z

x y z

.

------------------------Ht------------------------

THY HONG XUN VNH 53


14

THI TUYN SINH VO LP 10 CHUYN NG THP NM HC 2014-2015

Mn: Ton Thi gian lm bi: 150 pht. Ngy thi: 10/6/2014

Bi 1: Cho biu thcx 2 x 2 x 1

.x 1 x 1 x 2

a) Tm iu kin biu thc A c xc nh v rt gn A b) Tm gi tr ca x A t gi tr nh nht.

Bi 2: a) Tm cc gi tr x tha mn4

x 4x

b) Bc n trng cy trn mnh vn hnh ch nht, Bc d nh trng theo tng hng v mi hng c s cy bng nhau. Nu tng thm 1 hng nhng mi hng bt i 1 cy th s cy phi trng tng thm 10 cy. Nu bt i 1 hng nhng tng thm mi hng 2 cy th s cy phi trng tng thm 9 cy. Hi s lng cy m Bc n d nh trng l bao nhiu ?

Bi 3: a) Tm gi tr tham s m phng trnh x2+2mx2=0 c hai nghim phn bit x1, x2 tha x21+x22=5.

b) Mt hnh ch nht c chu vi bng 24 (mt), c di ng cho l 4 5 (mt). Hy tnh di cc cnh hnh ch nht .

Bi 4: Cho tam gic ABC vung ti A, k ng trung tuyn AM v ng cao AH. Gi D, E ln lt l hnh chiu ca H trn AB, AC. a) Chng minh DE2 = BH. HC v DE vung gc vi AM. b) Gi s din tch tam gic ABC bng 2 ln din tch t gic AEHD. Chng minh tam gic ABC vung cn. Bi 5: Cho tam gic ABC nhn ni tip ng trn (O), BH v CK l cc ng cao. Cc tip tuyn vi ng trn (O) ti B v C ct nhau ti S, cc ng thng BC v OS ct nhau ti M. a) Chng minh MB = MH. b) Chng minh rng AB. MH = AH. BS

c) Chng minh AHM ABS

15

S GIO DC &O TO C MAU K THI TUYN SINH LP 10 THPT

Nm hoc: 2014 2015

CHNH THC Mn thi : TON

Ngay thi 23/6/2014

Thi gian: 120 pht (khng k thi gian giao )

Bi 1 : (1,5 im)

a) Gii phng trnh 6x2 5x 6 = 0

b) Tm tham s m phng trnh :x2 +2(m +1)x +2m2 +2m +1 = 0 v nghim

THY HONG XUN VNH 54


Bi 2: (1,5 im)

a) Tnh gi tr ca biu thc A =1 1

6 2 6 2

b) Rt gn biu thc B = 1 2 2 1 2x x x vi 2 3x

Bi 3 :(2,0 im)

a) Gii h phng trnh: 2

8x y 6

x y 6

b) V th ca 2 hm s : y = x2 v y = 5x 6 trn cung h trc ta Oxy v tm ta giao im ca

hai th trn.

Bi 4:(2,0 im)

Mt hnh ch nht c chiu di gp 3 ln chiu rng. Nu c chiu di v chiu rng cung tng thm 5 cm

th dc mt hnh ch nht mi c din tch bng 153 cm2.Tm chiu di v chiu rng ca hnh ch nht ban u..

Bi 5: (3,0 im)

Cho tam gic ABC c 3 gc nhn, ni tip trong ng trn (O).Cc ng cao BF,CK ca tam gic ABC

ln lt ct (O) ti D,E.

a) Chng minh : T gic BCFK l t gic ni tip. b) Chng minh : DE //FK. c) Gi P,Q ln lt l im i xng vi B,C qua O.Chng minh ng trn ngoi tip tam gic AFK c

bn knh khng i khi A thay i trn cung nh PQ (khng trung vi cc im P,Q)

Ht..

Bi 1:

a) 26 5 6 0x x

25 4.6.6 25 144 169

5 13 3 5 13 2

12 2 12 3x hay x

b)Phng trnh :x2 +2(m +1)x +2m2 +2m +1 = 0 (a= 1;b=2(m+1);c=2m2 +2m+1)

' = (m+1)2 -2m2 -2m-1= m2 +2m+1-2m2 -2m-1= -m2 < 0 vi mi m

Vy phng trnh trn v nghim vi mi m m R

Bi 2:

a) A =1 1

6 2 6 2

6 2 6 2 2 6 6

6 4 26 2 6 2

THY HONG XUN VNH 55


b) B = 1 2 2 1 2x x x (vi 2 3x )

2

2 1 1 2 2 1 1 2B x x x x

2 1 1 2 2B x x (V 2

THY HONG XUN VNH 56


S GIO DC V O TO K THI TUYN SINH VO LP 10 THPT

TNH B RA-VNG TU Nm hc 2014 2015

CHNH THC MN THI: TON

Ngy thi: 25 thng 6 nm 2014

Thi gian lm bi: 120 pht (khng k thi gian giao )

Bi 1: (3,0 im)

a) Gii phng trnh: x2+8x+7=0

b) Gii h phng trnh: 3 5

2 4

x y

x y

c) Cho biu thc : 26

(2 3) 752 3

M

d) Tm tt c cc cp s nguyn dng (x;y) tho mn 4x2=3+y2 Bi 2: (2.0 im)

Cho parabol (P): 22y x v ng thng (D): y=x-m+1( vi m l tham s).

a) V Parabol (P) b) Tm tt c cc gi tr ca m (P)ct (D) c ng mt im chung. c) Tm ta cc dim thuc (P) c honh bng hai ln tung .

Bi 3: (1 im)

Hng ng phong tro V bin o Trng Sa mt i tu d nh ch 280 tn hng ra o. Nhng khi

chun b khi hnh th s hng ha d tng thm 6 tn so vi d nh. V vy i tu phi b sung thm 1 tu v

mi tu ch t hn d nh 2 tn hng. Hi khi d nh i tu c bao nhiu chic tu, bit cc tu ch s tn

hng bng nhau?

Bi 4: (3,5 im)

Cho ng trn (O) v mt im A c nh nm ngoi (O). K tip tuyn AB, AC vi (O) ( B,C l cc tip

im). Gi M l mt im di ng trn cung nh BC( M khc B v C). ng thng AM ct (O) ti im th 2 l

N. Gi E l trung im ca MN.

a) Chng minh 4 im A,B,O,E cng thuc mt ng trn. Xc nh tm ca ng trn .

b) Chng minh 2 180oBNC BAC c) Chng minh AC2=AM.AN v MN2=4(AE2-AC2). d) Gi I, J ln lt l hnh chiu ca M trn cnh AB, AC. Xc nh v tr cu M sao cho tch MI.MJ t gi tr

ln nht. Bi 5: (0,5 im)

Cho hai s dng x, y tha xy=3. Tm gi tr nh nht ca biu thc P=3 9 26

3x y x y

-------HT-------

THY HONG XUN VNH 57


ap an:

Bi 1:

1. Gii phng trinh v h PT

a) x2 +8x +7 = 0 Ta c: a-b+c=1-8+7=0 nn pt c hai nghim phn bit:

x1=-1; x2=-7

Vy tp nghim ca PT l : S={-1;-7}

b) 3 5 1 1

2 4 2 4 2

x y x x

x y y y

c) 6

(2 3) 75 6(2 3) 2 3 5 3 142 3

M

d) Ta c: 4x2-y2=3(2x+y)(2x-y)=3

2 3 1( )

2 1 1

2 1 1( )

2 3 1

2 1 1( )

2 3 1

2 3 1( )

2 1 1

x y xn

x y y

x y xl

x y y

x y xl

x y y

x y xl

x y y

Vy nghim dng ca pt l (1; 1)

Bi 2:

a) V th hm s:

x -2 -1 0 1 2

y=22x 8 2 0 2 8

b) Xt phng trnh honh giao im c (P) v (D): 22x = 1x m 2x2-x+m-1=0

=(-1)2-4.2(m-1)=9-8m

(P) v (D) c mt im chung th : =09-8m=0m=9

8

Vy vi m=9

8 th (P) v (D) c mt im chung.

c) im thc (P) m honh bng hai ln tung ngha l x=2y nn ta c:

THY HONG XUN VNH 58


y=2(2y)2y=8y2

0

1

8

y

y

Vy im thuc (P) m honh bng hai ln tung l (0;0) , (1

4,1

8)

Bi 3:

Gi x(chic) s tu d nh ca i( xN*, x

THY HONG XUN VNH 59


Bi 4:

a) Ta c: EM=EN(gt)OEMN 90oAEO

M 090ABO (AB l tip tuyn (O))

Suy ra: hai im B, E thuc ng trn ng knh AO. Hay A,B,E,O cng thuc mt ng

trn, tm ca ng trn l trung im ca AO.

b) Ta c: 2BOC BNC (gc tm v gc nt cng chn mt cung).

Mt khc: 0180BOC BAC

suy ra: 2 180oBNC BAC

(pcm)

c)

Xt AMC v ACN c

1( )

2

NAC chung

MCA CNA sdCM

AMC ACN(g.g)

2 .AM AC

AC AM ANAC AN

(pcm)

Ta c: AE2=AO2-OE2(p dng L Pi-ta-go vo AEO )

AC2=AO2-OC2(p dng L Pi-ta-go vo ACO )

Suy ra: AE2- AC2=OC2-OE2=ON2-OE2=EN2=

2 2

2 4

MN MN

hay MN2=4(AE2- AC2)

d) K MKBC, on AO (O) ={F}, AO BC ={H}

Ta c: MJK MCK ( t gic MJCK nt)

MCK MBI (cng chc cung MC)

MBI MKI (t gic MKBI nt)

Suy ra: MJK MKI (1)

Chng minh tng t ta cng c: MIK MKJ (2)

T (1) v (2) suy ra: MIK MKJ (g.g) 2 .MI MK

MK MI NJMK MJ

THY HONG XUN VNH 60


MI.MJ ln nht th MK phi ln nht. Mt khc M thuc cung nh BC nn MKFH vy MK

ln nht khi MK=FH. Hay M F Vy khi A, M, O thng hng th MI.MJ t gi tr ln nht.

Bi 5:

Ap dng bt Cosi ta c: 3 9

x y 2

276

xy (1)

3x+y2 3 6xy 26 13 26 13

3 3 3 3x y x y

(2)

T (1) v (2) suy ra:P= 3 9 26

3x y x y

6

13

3 P=

3 9 26

3x y x y

5

3

Vy MinP=5

3khi

3 1( 0)

3 3

x y x x

xy y

THY HONG XUN VNH 61


17

THY HONG XUN VNH 62


Gii S Lc

THY HONG XUN VNH 63


THY HONG XUN VNH 64


Q

P

MH

O

CB

A

THY HONG XUN VNH 65


18

S GIO DC V O TO K THI TUYN SINH VO LP 10 TRUNG HC PH THNG

BNH NH NM HC 2014 2015

d b Mn thi: TON Ngy thi: 28/6/2014

Thi gian lm bi: 120 pht (khng k thi gian pht )

----------------------------------------

Bi 1: (2,5 im)

a) Gii phng trnh: 3x 5 = x + 1

b) Gii phng trnh: 2 6 0x x

c) Gii h phng trnh: 2 8

1

x y

x y

d) Rt gn biu thc: P = 5

2 55 2

Bi 2: (1,5 im)

Cho phng trnh: 2 2 1 3 0 1x m x m

a) Chng minh phng trnh (1) lun c hai nghim phn bit vi mi gi tr m. b) Tm gi tr ca m phng trnh (1) c hai nghim i nhau.

Bi 3: (2,0 im)

Hai i cng nhn cng lm chung mt cng vic th hon thnh sau 12 gi, nu

lm ring th thi gian hon thnh cng vic ca i th hai t hn i th nht l 7 gi.

Hi nu lm ring th thi gian mi i hon thnh cng vic l bao nhiu?

Bi 4: (3,0 im) Cho ng trn tm O ng knh AB, trn cng mt na ng trn

(O) ly 2 im G v E (theo th t A, G, E, B) sao cho tia EG ct tia BA ti D. ng thng

vung gc vi BD ti D ct BE ti C, ng thng CA ct (O) ti im th hai l F.

a) Chng minh t gic DFBC ni tip. b) Chng minh: BF = BG

c) Chng minh: .

.

DA DG DE

BA BE BC

Bi 5: (1,0 im)Cho A = 1 1 1 1

....1 2 2 3 3 4 120 121

THY HONG XUN VNH 66


B = 1 1

1 ....2 35

Chng minh rng: B > A

BI GII

Bi 1: (2,5 im)

a) 3x 5 = x + 1 3x

b) 2 6 0x x Gii ra c nghim: 1 23; 2x x

c) 2 8

1

x y

x y

3 9 3

1 2

y y

x y x

d) P = 5

2 55 2

=

5 5 2

2 5 5 2 5 2 5 55 2 5 2

Bi 2: (1,5 im)

a) Phng trnh (1) c:

2

22 2 3 7' ' 1 3 3 4 02 4

b ac m m m m m m

, (v

23

0,2

m m

)

Vy: phng trnh (1) lun c hai nghim phn bit vi mi gi tr ca m.

b) PT (1) c hai nghim i nhau 2 1 00 1

10 33 0

mS mm

P mm

Vy vi m = 1 th phng trnh (1) c hai nghim i nhau.

Bi 3: (2,0 im)

Gi x (gi) l thi gian mt mnh i mt lm hon thnh cng vic. K: x > 12.

Thi gian mt mnh i hai lm xong cng vic l: x 7 (gi)

Trong 1 gi: + i mt lm c: 1

x (CV)

+ i hai lm c: 1

7x (CV)

+ C hai i lm c: 1

12(CV)

THY HONG XUN VNH 67


Ta c: PT: 21 1 1

31 84 07 12

x xx x

Gii PT c nghim: 1 228 ; 3x TM x KTM

Vy: i mt lm mt mnh sau 28 gi xong cng vic

i hai lm mt mnh sau 21 gi xong cng vic

Bi 4: (3,0 im)

a) Chng minh t gic DFBC ni tip.

Ta c: 0AF 90B (gc nt chn na ng trn)

Ta c: 090CDB CFB

t gic DFBC ni tip ng trn ng knh BC

b) Chng minh: BF = BG

Ta c: 090AEB (gc nt chn na ng trn)

090AEC

Ta c: 0180AEC ADC

T gic ADCE ni tip ng trn ng knh AC

1 1E C (v nt cng chn cung DA)

Ta c: 1 1B C (v nt cng chn cung DF ca ng trn ng knh BC)

Do : 1 1 AFE B AG BF BG BF BG

c) Chng minh: .

.

DA DG DE

BA BE BC

Ta chng minh c:

2

1

1

1

DO

F

E

G

BA

C

2

1

1

1

DO

F

E

G

BA

C

THY HONG XUN VNH 68


DGB DAE (g g) . .DG DB

DG DE DA DBDA DE

(1)

BEA BDC (g g) . .BE BA

BE BC BA BDBD BC

(2)

T (1) v (2) suy ra: . .

. .

DG DE DA DB DA

BE BC BA BD BA (pcm)

Bi 5: (1,0 im)

Ta c: A = 1 1 1 1

....1 2 2 3 3 4 120 121

=

=

1 2 2 3 120 121....

1 2 1 2 2 3 2 3 120 121 120 121

= 1 2 2 3 120 121

....1 1 1

= 2 1 3 2 ....... 121 120 = - 1 + 11 = 10 (1)

Vi mi k *,N ta c: 1 2 2 2 11

k kk k k k k

Do : B = 1 1

1 ....2 35

2 1 2 2 3 3 4 ..... 35 36B = 2 1 36 2 1 6 10 (2)

T (1) v (2) suy ra: B > A

GV: V Mng Trnh TRNG THCS CT MINH - PH CAT BNH NH

THY HONG XUN VNH 69


19

S GIO DC V O TO K THI TUYN SINH LP 10 THPT

TP.HCM Nm hoc: 2014 2015

CHNH THC MN: TON


Bai 1: (2 im)

Gii cc phng trnh v h phng trnh sau:

a) 2 7 12 0 x x

b) 2 ( 2 1) 2 0 x x

c) 4 29 20 0 x x

d) 3 2 4

4 3 5

x y

x y

Bai 2: (1,5 im)

a) V th (P) ca hm s 2y x v ng thng (D): 2 3 y x trn cng

mt h trc to .

b) Tm to cc giao im ca (P) v (D) cu trn bng php tnh.

Bai 3: (1,5 im)

Thu gn cc biu thc sau:

THY HONG XUN VNH 70


5 5 5 3 5

5 2 5 1 3 5

A

1 2 6

: 13 3 3

xB

x x x x x x (x>0)

Bai 4: (1,5 im)

Cho phng trnh 2 1 0 x mx (1) (x l n s)

a) Chng minh phng trnh (1) lun c 2 nghim tri du

b) Gi x1, x2 l cc nghim ca phng trnh (1):

Tnh gi tr ca biu thc : 2 21 1 2 2

1 2

1 1

x x x xP

x x

Bai 5: (3,5 im)

Cho tam gic ABC c ba gc nhn, ni tip ng trn tm O (AB < AC).

Cc ng cao AD v CF ca tam gic ABC ct nhau ti H.

a) Chng minh t gic BFHD ni tip. Suy ra 0AHC 180 ABC b) Gi M l im bt k trn cung nh BC ca ng trn (O) (M khc B

v C) v N l im i xng ca M qua AC. Chng minh t gic AHCN ni tip.

c) Gi I l giao im ca AM v HC; J l giao im ca AC v HN.

Chng minh AJI ANC

d) Chng minh rng : OA vung gc vi IJ

BI GII

Bai 1: (2 im)

Gii cc phng trnh v h phng trnh sau:

a) 2 7 12 0 x x

THY HONG XUN VNH 71


27 4.12 1

7 1 7 14 3

2 2

x hay x

b) 2 ( 2 1) 2 0 x x

Phng trnh c : a + b + c = 0 nn c 2 nghim l :

1 2 c

x hay xa

c) 4 29 20 0 x x

t u = x2 0 pt thnh :

2 9 20 0 ( 4)( 5) 0 u u u u 4 5 u hay u

Do pt 2 24 5 2 5 x hay x x hay x

d) 3 2 4

4 3 5

x y

x y

12 8 16

12 9 15

x y

x y

1

2

y

x

Bai 2:

a) th:

Lu y: (P) i qua O(0;0), 1;1 , 2;4

(D) i qua 1;1 , 3;9

b) PT honh giao im ca (P) v (D) l

THY HONG XUN VNH 72


2 2 3 x x 2 2 3 0 x x 1 3 x hay x (a-b+c=0)

y(-1) = 1, y(3) = 9

Vy to giao im ca (P) v (D) l 1;1 , 3;9

Bai 3:Thu gn cc biu thc sau

5 5 5 3 5

5 2 5 1 3 5

A

(5 5)( 5 2) 5( 5 1) 3 5(3 5)

( 5 2)( 5 2) ( 5 1)( 5 1) (3 5)(3 5)

5 5 9 5 15 5 5 9 5 153 5 5 3 5 5

4 4 4

3 5 5 5 2 5 5

1 2 6

: 13 3 3

xB

x x x x x x (x>0)

1 2 6:

3 3 ( 3)

1 ( 2)( 3) 6:

3 ( 3)

( 1). 1

x x

x x x x x

x x x

x x x

xx

x x

Cu 4:

Cho phng trnh 2 1 0 x mx (1) (x l n s)

a) Chng minh phng trnh (1) lun c 2 nghim tri du Ta c a.c = -1 < 0 , vi mi m nn phng trnh (1) lun c 2 nghim tri du vi

mi m.

b) Gi x1, x2 l cc nghim ca phng trnh (1):

Tnh gi tr ca biu thc :

THY HONG XUN VNH 73


2 21 1 2 2

1 2

1 1

x x x xP

x x Ta c 21 1x mx 1 v

2

2 2x mx 1 (do x1, x2 tha 1)

Do 1 1 2 2 1 2

1 2 1 2

mx 1 x 1 mx 1 x 1 (m 1)x (m 1)xP 0

x x x x

(V 1 2x .x 0 )

Cu 5

a) Ta c t gic BFHD ni tip do c 2 gc i

F v D vung 0180 FHD AHC ABC

b) ABC AMC cng chn cung AC

m ANC AMC do M, N i xng

Vy ta c AHC v ANC b nhau

t gic AHCN ni tip

c) Ta s chng minh t gic AHIJ ni tip

Ta c NAC MAC do MN i xng qua AC m NAC CHN (do AHCN ni tip)

IAJ IHJ t gic HIJA ni tip.

AJI b vi AHI m ANC b vi AHI (do AHCN ni tip)

AJI ANC

Cch 2 :

Ta s chng minh IJCM ni tip

Ta c AMJ = ANJ do AN v AM i xng qua AC.

M ACH = ANH (AHCN ni tip) vy ICJ = IMJ

IJCM ni tip AJI AMC ANC

B

A

F

C

O

D

K

H

M

x

I

J

Q

N

THY HONG XUN VNH 74


d) K OA ct ng trn (O) ti K v IJ ti Q ta c AJQ = AKC

v AKC = AMC (cng chn cung AC), vy AKC = AMC = ANC

Xt hai tam gic AQJ v AKC :

Tam gic AKC vung ti C (v chn na vng trn ) 2 tam gic trn ng dng

Vy 0Q 90 . Hay AO vung gc vi IJ

Cch 2 : K thm tip tuyn Ax vi vng trn (O) ta c xAC = AMC

m AMC = AJI do chng minh trn vy ta c xAC = AJQ JQ song song Ax

vy IJ vung gc AO (do Ax vung gc vi AO)

20

S GIO DC O TO

NINH THUN


NM HC 2014 2015

Kha ngy: 23 6 2014

Mn thi: TON


Bi 1: (2,0 im)

a) Gii phng trnh bc hai: x2 2x 2 = 0

THY HONG XUN VNH 75


b) Gii h phng trnh bc nht hai n:

25)(2

23

xyx

yx

Bi 2: (2,0 im) Cho hm s: y = 2x 5 c th l ng thng (d)

a) Gi A, B ln lt l giao im ca (d) vi cc trc ta Ox,Oy. Tnh ta cc im A, B v v ng thng (d) trong mt phng ta Oxy

b) Tnh din tch ca tam gic AOB Bi 3: (2,0 im)

Cho biu thc: P = 2222

33

.yx

yx

yxyx

yx

, x y

a) Rt gn biu thc P.

b) Tnh gi tr ca P khi: x = 347 v y = 324

Bi 4: (4,0 im)

Cho hnh ch nht ABCD c cnh AB = a ni tip trong ng trn tm O,

bn knh R (0 < a < 2R).

a) Tnh din tch ca hnh ch nht ABCD theo a v R.

b) Xc nh gi tr ca a theo R hnh ch nht ABCD c din tch

ln nht.

c) Mt ng thng d i qua O ct cc cnh AB, CD ln lt ti M, N v ct cc cnh AD, BC ko di ln lt ti P, Q. Chng minh rng: APM = CQN.

--------HT-------

HNG DN GII:

Bi 1: (2,0 im)

a) Gii phng trnh bc hai: x2 2x 2 = 0 = 3

1 1 3x ; 1 1 3x

THY HONG XUN VNH 76


b)

4

2

2

4

223

23

25)(2

23

y

x

x

y

yx

yx

xyx

yx

Bi 2: (2,0 im)

a) V (d) :

x 0 5/2

y = 2x 5 - 5 0

A(5/2;0)

B(0;-5); T v th

b) SA0B =1

2. OA.OB =

1

2.5.

5 25

2 4 (vdt)

Bi 3: (2,0 im)

a) Rt gn biu thc P.

P = 2222

33

.yx

yx

yxyx

yx

, vi x y

= ))((

.))((

22

22

yxyx

yx

yxyx

yxyxyx

=

yx

yx

b) P = yx

yx

x = 347 = 2 - 3 v y = 324 = 3 - 1

Vy: P = 2 - 3 3 1 1 3 2 3

3(2 3) ( 3 1) 3 2 3

Bi 4: (4,0 im)

a) Tnh din tch ca hnh ch nht ABCD theo a v R.

Ta c: SABCD = AB.BC = a. 22 ABAC = a. 224 aR

THY HONG XUN VNH 77


b) Xc nh gi tr ca a theo R hnh ch nht ABCD c din tch ln

nht.

V: 0 < a < 2R, nn: 2R a > 0

Ta c: 2222222 42)4(0)4( aRaaRaaRa

222 24 RaRa

Hay : SABCD 22R

Du = xy ra khi: a = 24 22 RaaR

Vy: Max SABCD = 2R2 khi: 2Ra

c) Chng minh rng: APM = CQN

- Trc ht, ta chng minh: AOM = CON (g.c.g) suy ra: AM = CN

- Xt APM v CQN c: AM = CN (cmt)

090 BA

AMP QNC (slt)

CQNAPM (g.c.g)

21

N

M

0

Q

B

C

A

D

P

THY HONG XUN VNH 78


S GIO DC V O TO HNG

YN

CHNH THC


NM HC 2014-2015

Mn thi : Ton

Thi gian lm bi 120 pht Ngy thi 23/6/2014

Cu 1: (2,0 im)

1) Rt gn biu thc: 2 8 2 3 2 6P 2) Tm m dng thng y = (m +2)x + m song song vi ng thng y = 3x 2

3) Tm honh ca im A trn parabol y = 2x2 , bit tung y = 18

Cu 2: ( 2,0 im). Cho phng trnh 2 2 3 0x x m ( m l tham s)

1) Tim m phng trnh c nghim x = 3. Tm nghim cn li.

2) Tm m phng trnh c hai nghim phn bit x1 , x2 tha mn : 3 3

1 2 8x x .

Cu 3: (2,0 im)

1) Gii h phng trnh : 2 3

3 2 1

x y

x y

2) Mt mnh vn hnh ch nht c chiu di hn chiu rng 12m . Nu tng chiu di thm 12m

v chiu rng thm 2m th din tch mnh vn tng gp i. Tnh chiu di v chiu rng ca

mnh vn .

Cu 4 ( 3,0 im) . Cho ABC c ba gc nhn ni tip trong ng trn tm O, bn knh R. H cc

ng cao AH, BK ca tam gic. Cc tia AH, BK ln lt ct (O) ti cc im th hai l D, E.

a) Chng minh t gic ABHK ni tip ng trn . Xc nh tm ng trn .

b) Chng minh : HK // DE.

c) Cho (O) v dy AB c nh, im C di chuyn trn (O) sao cho tam gic ABC c ba gc nhn.

Chng minh rng di bn knh ng trn ngoi tip CHK khng i.

Cu 5 ( 1,0 im). Gii h phng trnh

2 2

22

2 3 2 4 0

5 2 2 5

x y xy x y

x x y

THY HONG XUN VNH 79


S GIO DC V O TO HNG

YN

HNG DN


NM HC 2014-2015

Mn thi : Ton

Thi gian lm bi 120 pht

Ngy thi 23/6/2014

Cu 1 ( 2 im )

1) Rt gn : P 2( 8 2 3) 2 6

P 16 2 6 2 6 4

2) Tm m ng thng y = (m+2)x+m song song vi ng thng y = 3x 2. Hai ng thng song song khi v ch khi m+2 = 3 v m -2 . Do m = 1.

3) Tm honh ca im A trn parabol y = 2 x2 ,bit A c tung y = 18.

A

2

A A

y 18

y 2x

Suy ra xA = 3 .

Cu 2: ( 2,0 im). Cho phng trnh 2 2 3 0x x m ( m l tham s) (1)

1) Thay x = 3 vo phng trnh (1) ta c:

23 2.3 3 0 6 0 6m m m

Thay m = -6 vo PT (1) c dng: 2 2 3 0x x

Ta c: a b + c = 1+ 2 3 = 0

PT c hai nghim : x1 = -1

x2 = 3

Vy nghim cn li l x = -1

2) 2' 1 3 2m m

PT c hai nghim phn bit x1 , x2 ' 0 2 0 2m m

p dng nh l Viet ta c : 1 2

1 2

2

3

x x

x x m

THY HONG XUN VNH 80


Ta c:

3 3 2 2

1 2 1 2 1 1 2 2

2

1 2 1 2 1 2

8 8

3 8

x x x x x x x x

x x x x x x

Thay 1 2

1 2

2

3

x x

x x m

vo biu thc ta c

22 2 3 3 8 6 18 3m m m ( tha mn 2 m )

Vy m = - 3 phng trnh c hai nghim phn bit x1 , x2 tha mn : 3 3

1 2 8x x .

Cu 3: (2,0 im)

1) Gii h phng trnh :

2 3 4 2 6 7 7 1

3 2 1 3 2 1 3 2 1 1

x y x y x x

x y x y x y y

H PT cho c nghim ( x = 1; y = -1)

2) Gi chiu rng ca mnh vn hnh ch nht l x (m) K : x > 0

Th chiu di ca khu vn hnh ch nht l : x + 12 (m)

Din tch ca khu vn khi l: x(x + 12) ( m2)

Nu tng chiu di 12m v chiu rng ln 2m th :

Chiu di mi l : x + 12 + 12 = x + 24 (m)

Chiu rng mi l : x + 2 (m)

Din tch ca hnh ch nht mi l : ( x +2)( x + 24) (m2)

V din tch sau khi thay di gp i din tch ban u nn :

(x +2)( x + 24) = 2x( x+ 12)

x2 -2x 48 = 0

THY HONG XUN VNH 81


2'

1 2

1 1 48 49 7

1 7 1 78; 6

1 1x x

Vy chiu rng ca khu vn hnh ch nht l 8(m), chiu di ca khu vn l 20m

Cu 4 ( 3im )

Cho tam gic ABC nhn ni tip ng trn tm O bn knh R. Cc ng cao AH v BK ct ng trn ti im th hai theo th t D v E.

a) Chng minh t gic ABHK ni tip. Xc nh tm ca ng trn .

b) Chng minh DE//HK c) Cho (O;R) v dy AB c nh, im C di chuyn trn (O:R) sao

cho tam gic ABC nhn. Chng minh rng di ng trn ngoi tip tam gic CHK khng i.

a) T gic ABHK c OAKB AHB 90 . Suy ra T gic ABHK ni tip ng trn ng

knh AB.Tm O ca ng trn ny l trung im ca AB.

b) T gic ABHK ni tip nn KHA KBA . Xt (O)c KBA EDA . Suy ra

KHA EDA . Do HK//DE. c) Gi M l trung im ca AB M c nh OM khng i. Chng minh : AFBN l hnh bnh hnh suy ra F,M,N thng hng

Chng minh : CF = 2.OM khng i.

Chng minh CKFH ni tip ng trn ng knh CF. Suy ra di ng trn ngoi tip tam gic

CHK bng OM = CF

2 khng i

Cu 5 (1 im ) Gii h phng trnh :

2 2

2 2

x 2y 3xy 2x 4y 0 (1)

(x 5) 2x 2y 5 (2)

T (1) (x-2y) (x-y-2) = 0

xy

2

y x 2

*Xt x

y2

th (2) 2 2(x 5) x 5 . t x2 5 = a nn ta c h phng trnh :

F

N

M

D

EOK

H

C

BA

THY HONG XUN VNH 82


2

2

x 5 a

a x 5

suy ra x2 a2 -5= a-x 5 (a-x)(a+x+1) = 0

a x

a x 1

.

- Khi a = x ta c phng trnh x2 x 5 = 0

1,2 1,2

1 21x y 1 21

2

- Khi a = -x-1 th ta c phng trnh x2 + x 4 = 0

3,4 3,4

1 17 3 17x y

2 2

.

* Xt y = x-2 th (2) 2 2(x 5) 9

2 2

2

x 5 3 x 8 x 2 2 y 2 2 2

x 5 3 x 2 y 2 2

Vy h phng trnh cho c 8 nghim

22

UBND tNH BC NINH THI TUYN SINH VO LP 10 THPT CHUYN

S GIO DC V O TO NM HC 2014 - 2015

Mn Thi : Ton ( Dnh cho tt c th sinh )

Thi gian lm bi : 120 pht ( khng k thi gian giao )

Ngy thi : 20 thng 6 nm 2014

Cu I. ( 1, 5 im )

Cho phng trnh 06222 mmxx (1) , vi n x , tham s m .

1) Gii phng trnh (1) khi m = 1 2) Xc nh gi tr ca m phng trnh (1) c hai nghim x1 , x2 sao cho

2

2

2

1 xx nh nht.

Cu II. ( 1,5 im )

CHNH THC

THY HONG XUN VNH 83


Trong cng mt h to , gi (P ) l th ca hm s y = x2 v (d) l

th ca hm s

y = -x + 2

1) V cc th (P) v (d) . T , xc nh to giao im ca (P) v (d)

bng th .

2) Tm a v b th ca hm s y = ax + b song song vi (d) v ct (P)

ti im c honh bng -1

Cu III .( 2,0 im )

1) Mt ngi i xe p t a im A n a im B , qung ng AB

di 24 km . Khi i t B tr v A ngi tng vn tc thm 4km 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 .

2 ) Gii phng trnh 111 xxxx

Cu IV . ( 3,0 im )

Cho tam gic ABC c ba gc nhn v ba ng cao AA , BB ,CC ct

nhau ti H .V hnh bnh hnh BHCD . ng thng qua D v song song vi BC

ct ng thng AH ti M .

1) Chng minh rng nm im A, B ,C , D , M cng thuc mt ng trn. 2) Gi O l tm ng trn ngoi tip tam gic ABC .Chng minh rng BM

= CD v gc BAM = gc OAC .

3) Gi K l trung im ca BC , ng thng AK ct OH ti G . Chng minh

rng G l trng tm ca tam gic ABC.

Cu V .( 2, 0 im )

1) Tm gi tr nh nht ca biu thc P = a2 + ab + b2 3a 3b + 2014 . 2) C 6 thnh ph trong c 3 thnh ph bt k th c t nht 2 thnh ph

lin lc c vi nhau . Chng minh rng trong 6 thnh ph ni trn tn ti 3 thnh ph lin lc c vi nhau.

.................Ht...............

THY HONG XUN VNH 84


( ny gm c 01 trang)

H v tn th sinh :..................................................................S bo danh

:..........................................

Hng dn s lc thi mn ton dnh cho tt c th sinh nm hc

2014-2015

Thi vo THPT chuyn Tnh Bc Ninh v cu V chuyn ton

Cu I. ( 1, 5 im )

Cho phng trnh 06222 mmxx (1) , vi n x , tham s m .

1) Gii phng trnh (1) khi m = 1 2) Xc nh gi tr ca m phng trnh (1) c hai nghim x1 , x2 sao cho

2

2

2

1 xx nh nht.

HD :

1) GPT khi m =1

+ Thay m =1 v o (1) ta c x2 + 2x 8 = 0 ( x + 4 ) ( x 2 ) = 0 x = { -

4 ; 2 }

KL :

2) x t PT (1) : 06222 mmxx (1) , vi n x , tham s m . + Xt PT (1) c 05162 221' mmm

(lun ng ) vi mi m => PT (1) lun c hai nghim phn bit x1 ; x2 vi mi

m

+ Mt khc p dng h thc vit vo PT ( 1) ta c :

62

2

21

21

mxx

mxx (I)

+ Li theo v (I) c :A = x12 + x22

= ( x1 + x2 )2 2 x1x2

THY HONG XUN VNH 85


= ( - 2m )2 + 2 ( 2m + 6 )

= 4m2 + 4m + 12

= ( 2m + 1)2 + 11 11 vi mi m => Gi tr nh nht ca A l 11 khi m = 2

1 .

KL :

Cu II. ( 1,5 im )

Trong cng mt h to , gi (P ) l th ca hm s y = x2 v (d) l

th ca hm s

y = -x + 2

1) V cc th (P) v (d) . T , xc nh to giao im ca (P) v (d)

bng th .

2) Tm a v b th ca hm s y = ax + b song song vi (d) v ct (P)

ti im c honh bng -1

HD : 1) v ch nh xc v xc nh c giao i m ca (P) v (d) l M ( 1 ; 1) v

N ( -2 ; 4 )

2)T m c a = -1 v b = 0 =>PT ca l y = - x

Cu III .( 2,0 im )

1) Mt ngi i xe p t a im A n a im B , qung ng AB

di 24 km . Khi i t B tr v A ngi tng vn tc thm 4km 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 .

2 ) Gii phng trnh 111 xxxx

HD :

THY HONG XUN VNH 86


1) G i x ( km /h ) l v n t c ng i i xe p t A -> B ( x > 0 ) . L lun a ra PT :

2

1

4

2424

xx => x = 12 ( t/m ) . KL : ............

2) KX 10 x t 0 < a = xxa

xx

12

11

2

+ PT m i l : a + 12

12

a a2 + 2a 3 = 0

32 de va dap an ts 10 cac tinh thanh 1415

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