[vnmath.com]-dethi hoc ki 1 Toan 10-dap an (2010-2011)

8/8/2019 [vnmath.com]-dethi hoc ki 1 Toan 10-dap an (2010-2011)

1/16Trang 1

THI TH KIM TRA HC K IMn ton lp 10 nm hc 2010 - 2011

Thi gian 90 (Khng k thi gian giao )

Cu I: (2,0 im)

1). Cho tp hp { }7; 6; 5,...,8;9;10M =

Lit k cc phn t ca tp hp { }| 3 A x x M = .

2). Cho cc tp hp { }| 5 1 A x x=


2/16Trang 2

II 2.01 1.0

2 AB AD AC + =

(quy tc hnh bnh hnh) 0.5

2OC=

(O l trung im caAC) 0.5

2 1.0

( )1; 2 AM x y=

; ( ) ( )3;1 , 5; 2 AB BC = =

0.25

2 AM AB BC + =

( )1 2. 3 5

2 2.1 2

x

y

+ =

+ =

0.25

12

2

x

y

=

= . Kt lun: ( )12; 2M . 0.5

III 2.01 1.0

( )2 5 2. 1 5 3A Ay x= + = + = . Suy ra ( )1;3A . 0.25

( ) : 2d y x m= + i qua im ( )1;3A nn ta c 3 1 2m= + 0.5

Gii c 2m = 0.252 1.0

Ta im ( )0;1K 0.25

( ) 2: 2P y x bx c= + + i qua hai im ( ) ( )1; 1 , 0;1M K nn ta c h2

2

1 1 2 .1

1 0 2 .0

b c

b c

= + +

= + +

2 0

1

b c

c

+ =

=

0.5

32

1

b

c

=

=

. Kt lun:3

; 12

b c= = . 0.25

IV 2.0

1 2 2 2 2sin cos 1 cos 1 sin + = = 1.02

2 12 25cos 113 169

= =

0.25

Do gc nhn nn cos 0 > . Suy ra25 5

cos169 13

= = . 0.25

sin 12 5 12tan :

cos 13 13 5

= = = 0.25

2 22 2 12 5 1132sin 7 cos 2. 7.

13 13 169P

= = =

0.25

2 1.0

Gi ta ca Cl ( );0C x , x .

( ) ( )2;3 , 1; 1 BA BC x= =

0.25

ABC vung tiB AB BC . 0BA BC = 0.25

( ) ( ) ( )1 . 2 1 .3 0x + = 0.25

1

2x = . Kt lun:

1;0

2C

0.25

V 2.0

1 2 1 2x x = (1) 1.0

( )2

2 0

2 1 2

x

x x

= 0.25

http://www.vnmath.com


3/16Trang 3

2

2

6 5 0

x

x x

+ =

2

1

5

x

x

x

=

=

1x = 0.5

Tp nghim ca (1) l { }1T = . 0.25Hc sinh c thbin i h qu (Cn nu iu kin xc nh)!

2 1.0

Vi 3 5x ta c ( )( )3 5 0Q x x=

0 3Q x= = hoc 5x = .0.25

Vy[ ]3;5

min 0Q = 0.25

Vi 3 5x ta c 5 0x v 3 0x . p dng bt ng thc Csi ta c

( )( )3 5

3 52

x xQ x x

+ =

Hay 1Q .1 5 3 4Q x x x= = =

0.25

Vy[ ]3;5

max 1Q = 0.25

. THI THKIM TRA HC K IMn ton lp 10 nm hc 2010 - 2011

Thi gian 90 (Khng k thi gian giao )------------------------------------

A/Phn chung ( Gm 5 bi , bt buc cho mi hc sinh) :

Bi 1 (2 im): Cho hm s 2 2= + y x x c th (P)1) Lp bng bin thin v v th (P).2) T th (P), hy nu cch v v v th (P1) ca hm s

2 2= + y x x .

Bi 2 (1,5 im): Gii v bin lun theo tham s mphng trnh:1

1 2

x x m

x x

+ +=

+

Bi 3 (1,5 im): Cho tam gic ABC c trng tm G. D v E l hai im xc nh bi: AD AB=

v2

3EA EC =

.

1) Chng minh ( )1

3 AG AB AC = +

2) Chng minh ba im D, G, E thng hng.

Bi 4 (1,5 im): Trong mt phng vi h ta Oxy cho cc im A(6;2); B(-2;-2); C(3;8) .1) Chng minh tam gic ABC vung ti A. Tnh di trung tuyn i qua A ca tam gic ny.2) Tm im E t gic ABEC l hnh bnh hnh.

Bi 5 (1 im): Tm gi tr nh nht ca hm s: 1( ) 32

y f x xx

= = + ++

vi x>-2

B/Phn t chn ( Hc sinh chn mt trong hai phn sau) : Phn dnh cho ban nng cao( Gm 6A v 7A):

Bi 6A (1,5 im): Cho h phng trnh0

1

x my

mx y m

=

= +

1) Tm m h phng trnh c v s nghim.2) Vit tp hp nghim ca h phng trnh trong cu 1).



4/16Trang 4

Bi 7A (1 im): Cho hnh vung ABCD c cnh bng a. Mt ng trn c bn knh bng6

3

a

i qua hai nh A, C v ct cnh BC ti E (khng cn chng minh s duy nht ca im E).1) Tnh di on AE.

2) Tnh s o gcBAE

Phn dnh cho ban c bn ( Gm 6B v 7B):Bi 6B (1,5 im): Cho phng trnh 2 1 0 x x m+ + = .

1) Tm m phng trnh c mt nghim m v mt nghim dng.2) Tm m phng trnh c mt nghim m, mt nghim dng v tr stuyt i ca mt

trong hai nghim bng hai ln tr stuyt i ca nghim kia.

Bi 7B (1 im): Cho tam gic cn ABC c AB = AC = a v 0120BAC= . Tnh gi tr ca biu

thc: . . .T ABCB CBCA AC BA= + +

theo a ./.===============================================

P N BI KIM TRA HC K IMn ton lp 10 nm hc 2009 - 2010

Bi Cu Ni dung gii im

1 1) + nh I(-1;-1)+ Do a=1>0 nn c BBT:

x - -1 +

2 2= + y x x

+ +

-1

+Trc i xng x = - 1 th ct hai trc ta ti cc im O(0;0); A(-2;0) V i qua im

B(1;3)

+ th:

0.25

0.5

0.25

0.25

1.25

2) + C ( )22 2 2 , x x x x x R+ = + nn hm s 2 2= + y x x l hm s

chn. Suy ra th i xng qua trc trung (1)

+ C2 2

2 2 , 0 x x x x x+ = + , suy ra th (P1) v th (P) trng nhautrong min x khng m.(2)+T (1) v (2) suy ra th (P1)l:

0.25

0.25

0.25

0.75

8

6

4

2

-2

-10 -5 5

B

A

5

4

3

2

1

-1

y

-4 -2 2 4

OJJ

OI

IO

B

A



5/16Trang 5

2 11 2

x x m

x x

+ +=

+(1). /k:

1

2

x

x

. C (1) (m - 4)x = 2 + m (2)

+Nu m = 4 th (2): 0x = 6, pt v nghim

+Nu m 4 th (2) 2

4

mx

m

+=

.

+2

.

4

mx

m

+=

l nghim ca (1) khi v ch khi

21

4

2 24

m

m

mm

+

+

0. 6

23 6

mm

m

+Kt lun:-Nu m = 4 hoc 2m th (1) v nghim

-Nu m 4 v 2m th (1) c nghim duy nht2

4

mx

m

+=

0.25

0.25

0.25

0.25

0.25

0.25

1.5

3 1) +V ng hnh

+

( ) ( )2 2 1 1

3 3 2 3 AG AM AB AC AB AC

= = + = +

0.25

0.25

0.5

2)+ ( )

1 5 12

3 3 3 DG AG AD AB AC AB AB AC = = + = +

(1)

+2 2

2 25 5

DE AE AD AC AB AB AC = = = +

(2)

+T (1) v (2) suy ra6

5

DE DG=

. Vy ba im D, G, E thng hng

0.5

0.25

0.25

1

4 1) + ( 8; 4)AB =

; ( 3;6)AC=

.

+ Suy ra . 0AB AC=

. Suy ra tam gic ABC vung ti A.

+ Trung im M ca BC l M(1

;32

);11

( ;1)2

AM =

+ di trung truyn2

211 125 5 512 4 2

AM

= + = =

0.25

0.25

0.25

0.25

1

2) + ( 8; 4)AB =

;

( 3; 8)CE x y=

vi E(x;y)

+T gic ABEC l hnh bnh hnh khi v ch khi AB CE= 8 3 5

4 8 4

x x

y y

= =

= = .

Vy E(-5;4)

0.25

0.25

0.5

5+C ( )

1 1( ) 3 2 1

2 2 y f x x x

x x= = + + = + + +

+ +

+Do x>-2 nn x +2>0.

p dng bt ng thc C-Si cho hai s dng ( )2x + v1

2x +ta c

0.25

G

M

A

B

C

E

D



6/16Trang 6

( ) ( )1 1

2 1 2 2 . 1 2 1 32 2

x xx x

+ + + + + = + =+ +

.

+Du = xy ra khi v ch khi ( )1

22

xx

+ =+

( )2 1

2 13

xx

x

= + =

= .

(loi x = -3).

+Suy ra( 2; )

( ) ( 1) 3 Min f x f +

= = .

0.25

0.25

0.25

1

6A 1) +D = m2 - 1; Dx = m(m+1); Dy = m + 1+Mun h c v s nghim th D = 0 , suy ra (m = 1) V (m = -1)+ Vi m = -1 c Dx = Dy = 0. nn h phng trnh c VSN+ Vi m = 1 c Dx = Dy = 2 0. nn h phng trnh v nghim

Vy m = -1

0.250.250.250.25

0.25

1.25

2) +Vi m = -1, h phng trnh tr thnh x + y = 0. Tp nghim ca hphng trnh l: }{ RttytxyxS === ,,/);( 0.25 0.25

7A 1) + 045=ACEgc A D

B E C

+Tam gic AEC: 00

2 2 .sin 45sin45

AE R AE R= =

2 3

3

aAE =

0.25

0,25

0.25

0.75

2)

Tam gic vung ABE c 3

cos 22 33

AB aBAE AE a= = =

030BAE = 0.25 0.25

6B 1) Phng trnh c mt nghim m, mt nghim dng khi v ch khia.c m 1 0 1m= < <

0.25 0.25

2) +Vi 1m < (*), phng trnh c mt nghim m, mt nghim dng.+Gi hai nghim ny l x1, x2 v gi s/x1/=2/x2/

2 21 24x x =

+Kt hp Vi-t ta c

=

=+

=

1

1

4

21

21

22

21

mxx

xx

xx

22 2

1 2

22 2

1 0(1)

1

3 2 1 0(2)

x x m

x x

x x

+ + =

= =

+Gii (2) c 2 21

1;3

x x= = .

Ln lt th vo (1) tm c 1;m = 119

m = (Loi do (*))

Vy 1m =

0.250.25

0.25

0.5

1.25



7/16Trang 7

7B+ 2 0 2

3. cos30

2 AB CB a a= =

+ 2 0 23

. 3 cos302

CB CA a a= =

+ 2 0 21

. cos602

AC BA a a= =

Vy 212

T a=

0.25

0.25

0.25

0.25

1


Thi gian 90 (Khng k thi gian giao )---------------------------

I. PHN CHUNG CHO TT C HC SINH ( 7 im ).

Cu I ( 1 im ). Xc nh tp hp sau v biu din kt qu trn trc s: ( - 1; 7 ) \ [ 2; 3 ]

Cu II ( 2 im ).

1. Xc nh cc h s a, b ca parabol y = ax2 + bx 3 bit rng parabol i qua im

A ( 5; - 8 ) v c trc i xng x = 2.

2. V th hm s y = - x2 + 4x 3.

Cu III ( 2 im ).

1. Gii phng trnh: 3x22x =+ .

2. Gii v bin lun phng trnh m2 x 3 = 9x + m theo tham s m.

Cu IV ( 2 im ).

1. Cho t gic ABCD. Gi M, N ln lt l trung im ca hai ng cho AC v BD.

Chng minh:

=+ MNCDAB .2

2. Trong mt phng Oxy, cho hai im A ( - 1; 0 ), B ( 2; 3 ). Tm ta im N trn

trc tung sao cho N cch u hai im A v B.

II. PHN RING ( 3 im ). Hc sinh chc chn mt trong hai cu Va hoc Vb

Cu Va. ( cbn)

1. Xt tnh chn, l ca hm s: f ( x ) = x22x +

2. Ba bn An, Bnh, Chi

i mua tri cy. B

n An mua 5 qu

cam, 2 qu

qut v 8 qu

to vi gi tin 95000 ng. Bn Bnh mua 1 qu cam, 5 qu qut v 1 qu to vi

gi tin 28000 ng. Bn Chi mua 4 qu cam, 3 qu qut v 2 qu to vi gi tin

45000 ng. Hi gi tin mi qu cam, qut, to.

3. Cho cosa =51

. Tnh gi tr ca biu thc P = 3.sin2a + 2.cos2a.

Cu Vb. ( nng cao)

1. Xt tnh ng bin, nghch bin ca hm s: f( x) = x2 2x + 3 trn khong ( 1; + )

120

A

B C



8/16Trang 8

2. Chng minh rng, vi 3 s a, b, c dng ta c: abcca

cb

c

ba

b

a8

+

+

+

3. Cho sina =5

1( 900 a 1800 ). Tnh cosa v tana

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

P N TON 10. HC K I . 2010 2011

Cu Ni dung imI 1,0 im.

+ ( - 1; 7 ) \ [ 2; 3 ] = ( - 1; 2 ) ( 3; 7 )+ Biu din kt qung, c ch thch

0,50,5

II 2,0 imII. 1 1,0 im

+ T gi thit ta c h PT:

=

+=

22

35258

a

bba

=

=

4

1

b

a

+ KL

0,5

0,25

0.25

II. 2 1,0 im+ nh I ( 2; 1 ), trc i xng x = 2, b lm quay xung+ Lp bng gi tr ( c giao im ca th vi 2 trc ta )+ Vng th

0,250,250,5

III 2,0 imIII. 1 1,0 im

+ k: x - 1+ Bnh phng 2 v ta c PT h qu: 2x + 2 = x2 6x + 9 x2 -8x + 7 = 0 x = 1 ( tha k ) hoc x = 7 ( tha k )+ Th li v kt lun PT c 1 nghim x = 7

0,250,25

0,250,25

III. 2 1,0 imPT ( m2 9 ) x = m + 3

+ Nu m 3 PT c nghim duy nht x =3

1m

+ m = 3 : PT v nghim, m = - 3 PT nghim ng vi mi x R+ KL

0,25

0,250,250,25

IV 2,0 imIV. 1 1,0 im

+

++= NBMNAMAB ( 1 ),

++= NDMNCMCD ( 2 )+ Cng ( 1 ) v ( 2 ), gii thch do M, N trung im, suy ra kt qu

0,50,5

IV. 2 1,0 im+ N oy suy ra N ( 0; y )+ NA = NB NA2 = NB2 y = 2+ KL

0,250,5

0,25Va 3,0 im

Va. 1 1,0 im+ Tp xc nh: D = [ - 2; 2 ], mi x D suy ra x D+ Chng minh f ( - x ) = - f ( x )

0,250,5



9/16Trang 9

+ KL: Vy hm s l trn D 0,25Va. 2 1,0 im

+ Gi x, y, z l gi tin mi qu cam, qut, to ( x,y,z > 0 )

+ t gt ta c h PT:

=++

=++

=++

45000234

280005

95000825

zyx

zyx

zyx

=

=

=

8000

3000

5000

z

y

x

+ KL

0,25

0,5

0,25

Va. 3 1,0 im

+ sin2a = 1 cos2a =25

24

+ P =25

74

0,5

0,5

Vb 3,0 imVb. 1 1,0 im

+ x1, x2 ( 1; + ), x1 x2 ,( )

21

21 )(

xx

xfxf

= x1 + x2 2

+ Gii thch c x1 + x2 2 > 0+ KL: hm sng bin trn ( 1; + )

0,5

0,250,25

Vb. 2 1,0 im

+ Bt ng thc C si cho hai s dngb

av a ta c:

b

aa

b

a 22+

Tng t c hai bt ng thc cn li+ Nhn ba bt ng thc v theo v suy ra pcm

0.5

0,5

Vb. 3 1,0 im

+ cos

2

a = 1 sin

2

a = 25

24

cosa = - 5

24

( v gc a t nn cosa < 0 )

+ tana =12

6

cos

sin=

a

a

0,5

0,5


Thi gian 90 (Khng k thi gian giao )

I/.PHN CHUNG: (7im) (Dnh cho tt c cc hc sinh)Cu 1: (2im)

1/.Cho hai tp hp [ )0;2 , (1;3)A B= = .Hy xc nh cc tp hp :, , \A B A B A B

2/.Kho st s bin thin v v th hm s : 2 4 5 y x x= + + Cu 2: (2im)

1/.Xt tnh chn l ca hm s: ( ) 1 1 f x x x= +

2/.Cho phng trnh : 2 22 0 x mx m m + = .Tm tham s m phng trnh c hai

nghim phn bit ,1 2x x tha mn :2 2 31 2 1 2 x x x x+ =

Cu 3: (3im)1/.Trong mt phng oxy cho: (1; 2), ( 3; 4), (5; 6) A B C



10/16Trang 10

a/.Chng minh ba im , ,A B Ckhng thng hng.b/.Tm ta trng tm G ca tam gic ABC.

2/.Cho3 0 0sin (0 90 )5

= < < .Tnh gi tr biu thc :1 t an1+tan

P

=

II/.PHN RING: (3im) (Hc sinh chn Cu4a hoc Cu 4b lm)Cu 4a: (3im) (Dnh cho hc sinh hc sch nng cao)

1/.Gii phng trnh : 2 24 9 6 4 9 12 20 0 x x x x + + =

2/.Tm m h phng trnh : 4

mx y m

x my

+ =

+ =c nghim duy nht l nghim

nguyn.

3/.Cho tam gic ABC vung cn ti A c 2 BC a= .Tnh : . , .CACB AB BC

Cu 4b: (3im) (Dnh cho hc sinh hc sch chun)

1/.Gii phng trnh: 4 27 12 0x x + =

2/.Gii h phng trnh:2 2 13

6

x y

xy

+ =

=

3/.Trong mt phng oxy cho tam gic ABC vi (1; 2), (5; 1), (3;2) A B C .

Tm ta im D t gic ABCD l hnh bnh hnh............ Ht..........

P N CHM MN TON KHI 10 HC K I NM HC 2010-2011.

Cu Ni dung im

1.1

[ )0;3A B = 0.25

(1;2)A B = 0.25

[ ]\ 0;1A B = 0.25

1.2

TX: D = ,ta nh (2;9)I 0.25

1a=

:Parabol quay b lm xung di v nhn 2x=

lm trc i xng. 0.25x 2 + y 9

0.25

0.5

2.1TX:D = , x D x D 0.25

( ) 1 1 f x x x = + 0.25

( ) 1 1 ( ) f x x x f x = + = 0.25

Kt lun: Hm s l 0.25

1 0

8

6

4

2

- 5 5 1 0

-1

5

y

O

9I

2



11/16Trang 11

2.2

/ 2 2 2( ) 0, 2 , .1 2 1 2m m m m S x x m P x x m m = = > = + = = = 0.25

2 2 23 ( ) 5 01 2 1 2 1 2 1 2x x x x x x x x+ = + =

2 24 5( ) 0

02 5 05

m m m

mm m

m

=

= + =

=

0.5

Kt lun : 5.m=

0.25

3.1a

( 4;2)AB = , AC (4;4)=

0.25

4 24 4

0.25

AB

khng cng phng vi AC

0.25

, ,A B Ckhng thng hng. 0.25

3.1b

13

x x xBA CxG

+ += =

0.25

43

y y yBA CyG

+ += =

0.25

Trng tm tam gic ABC l : (1;4)G 0.25

3.2

s25

3 9 40 0 2in ,(0 90 ) os = 1-sin 15 5

c = < < = = 0.25

3tan

4 =

0.25

3 11 tan 1

4 4 = =

0.25

3 71 tan 1

4 4+ = + =

0.25

1 tan 1

1 tan 7P

= =+ 0.25

4a.1

9 1112 24 9 12 (2 ) 0,4 16

x x x x + = + > 0.25

t : 24 9 12 0 y x x= + > ,phng trnh trv:22 6 8 04

yy y

y

= + =

= 0.25

2 22 4 9 12 2 4 9 8 0 y x x x x= + = + = : Phng trnh v nghim 0.25

24 4 9 12 49 14524 9 4 0

8 y x x x x x= + =

= = 0.25

4a.2

12 11

mD mm= = . Vi : 1m th h phng trnh c nghim duy nht v

1x = khng tha mn h phng trnh.Nn : 1x

0.25

T PT th nht ta c :1

ym

x=

thay vo PT th hai ta

0.25



12/16Trang 12

c: 2 25 (4 ) 0 x x y + =

25 9 422 25 (4 ) 0

25 9 42

yx

x x y

yx

+ +=

+ =

+=

x cn phi c

,2 29 4 ( 2 )( 2 ) 9,y n n n y n y y+ = + = 2 1

2 9

n y

n y

=

+ =

hoc2 1

2 9

n y

n y

=

+ =

hoc2 9

2 1

n y

n y

=

+ =hoc

2 9

2 1

n y

n y

=

+ = hoc

2 3

2 3

n y

n y

=

+ =hoc

2 3

2 3

n y

n y

=

+ = Gii ra

c : .2, 2,0y =

0.25

Th li :

2y = h c nghim : ( ) ( )0;2 , 5;2 2m = hoc1

.2

m =

2y = h c nghim : ( ) ( )0; 2 , 5; 2 2m = hoc12

m =

0y = h c nghim : ( ) ( )1;04;0 , 0m =

0.25

Vy : 1 12; ;0; ;22 2

m

0.25

4a.3

Tnh c : AB AC a= = 0.25

20 2. . . os45 . 2.2

CACB AC CB c a a a= = =

0.25

22

20. . . . os45 . 2. a AB BC BA BC BA BC c a a = = = =

0.25

4b.1

t : 2 0t x= a v phng trnh 2 7 12 0t t + = 0.25

Gii c : 34

tt

=

= 0.25

23 3 3t x x= = = 0.25

24 4 2t x x= = = .Kt lun phng trnh c 4 nghim : 3, 2x x= = 0.25

4b.2

2 2 2 2

5

613 ( ) 2 13 ( ) 25

6 6 6 5

6

x y

xy x y x y xy x y

xy xy xy x y

xy

+ =

=+ = + = + = = = = + = =

0.5

5

6

x y

xy

+ =

=

2

3

x

x

=

=hoc

3

2

x

y

=

=

0.25

5 2

6 3

x y x

xy y

+ = =

= = hoc

3

2

x

y

=

= 0.25

H phng trnh c 4 nghim : (2;3), (3;2), ( 2; 3), ( 3; 2) 0.25

4b.3

Gi ( ; )D x y , ( 1; 2), ( 2;3) AD x y BC = + =

0.5

T gic ABCD l hnh bnh hnh nn:1 2

2 3

x AD BC

y

= =

+ =

0.25



13/16Trang 13

Gii c :1

1

x

y

=

=.Kt lun : ( 1;1)D 0.25

Ch : Nu hc sinh lm cch khc m ng th ty theo gio vin chm cho cc phn imtng ng sao cho hp l.


Thi gian 90 (Khng k thi gian giao )I. PHN CHUNG: (7 im)

Cu 1: (1 im)

1. Cho A [ ]0;4= , B [ ]2;7= Xc nh tp ,A B A B

2. Tm tp xc nh ca hm s1

23

y xx

= +

Cu 2: (2 im)

1. Gii phng trnh 2 1 1x x = +

2. Gii v bin lun theo m phng trnh 2 2m x m x m+ = + Cu 3: (1,5 im) Cho hm s 2 4 y x x m= + , c th (P), m l tham s.

1. V th ca hm s khi m = 3.2. Vi gi tr no ca m th th (P) tip xc vi trc ox?

Cu 4: (2.5 im) Cho tam gic ABC bit nh A (0;-4), B(-5;6), C(3;2)1. Tm ta trng tm G, trc tm H ca tam gic ABC.2. Tm ta tm O ca ng trn ngoi tip tam gic ABC v chng minh:

2GH GO=

II. PHN RING : (3 im) (Hc sinh thuc ban no ch lm phn dnh ring cho ban )

Phn A: (Dnh cho hc sinh hc ban KHTN)Cu 5: (1 im) Cho a, b, c l cc s dng chng minh rng :

9

2

a b c a b c a b c

a b b c c a

+ + + + + ++ +

+ + +

Cu 6: (1 im) Gii h phng trnh:

1 2 1 22 1 1 2x y

x y + + = +

+ + = +

Cu 7: (1im) Cho tam gic ABC. Tm tp hp cc im M sao cho:

3 . MA MB MC MB MC + + = +

Phn B: (Dnh cho hc sinh hc ban cbn)Cu 5: (1 im) Cho x , y, z l cc s dng chng minh:

2

xy yz zx x y z

x y y z z x

+ ++ +

+ + +

Cu 6:(1 im) Gii h phng trnh sau ( khng s dng my tnh )

2 3 1

5 7 35 5 2

3 7 3

xy

x y

+ =

=

Cu 7: (1 im) Cho tam gic ABC. Gi I l trung im ca AB. Chng minh :2

2. .4

ABCACB CI=

(Ht)

P N V THANG IM THI HC K I MN TON LP 10NM HC 2010 - 2011



14/16Trang 14

p nThangim

Ghich

I. Phn chung: (07im)Cu 1: (01im)1. [ ]0;7A B =

[ ]2;4A B =

2. iu kin: 2 0 23 0 3x x

x x

>


15/16Trang 15

Gi ( );H HH x y , H l trc tm ABC. 0

. 0

AH BC AH BC

BH AC BH AC

=

=

( ) ( )

( ) ( )

( )

8 4 4 0

3 5 6 6 0

33;2

2

H H

H H

H

H

x y

x y

xH

y

+ + =

+ + =

=

=

2. (1,25)Do H C BC AC ABC vung ti C

O l trung im ca AB5

;12

O

Ta c:11 2 11 1

; , ;3 3 6 3

GH GO

= =

2GH GO =

0,25

0,5

0,25

0,5

0,25

0,25

0,25II.Phn ring: (03 im)Phn A:

Cu 5 (01im)

( )

( ) ( ) ( )

( )( )( )3

1 1 1

1 1 1 1=

2

13. .

2

a b c a b c a b ca b c

a b b c c a a b b c c a

a b b c c aa b b c c a

a b b c c a

+ + + + + + + + = + + + +

+ + + + + +

+ + + + + + +

+ + +

+ + + 31 1 1 9

3. . .2a b b c c a

=+ + +

(Phi ni c: p dng BT C-si cho 3 skhng m)

0,25

0,25

0,5

Cu 6: (01im)

iu kin:1 2

1 2

x

y

T h pt 1 1 2 2 x y x y + + =

01 1 2 2 1 1 2 2

x y y x x y x y

x y x y x y x y

x y

= + =

+ + + + + + + +

=

Thay x y= vo pt: 1 2 1 2x y+ + = + ta c:

( )( )0

1 2 21

0 0; 1 1

xx x

x

x y x y

=+ =

=

=

= =

=

Kt lun: H pt c hai nghim (0,0); (1,1)

0,25

0,25

0,25

0,25Cu 7: (01 im)Gi G l trng tm ca ABC , D l trung im ca BC ta c:

0,5



16/16

*Lu : Mi cch gii ng u cho im ti a.---------Ht---------

( )

2 3

2.3 3.2

*

MA MB MC MB MC

MG MD

MG MD MG MD

+ + = +

=

= =

T (*) suy ra M nm trn ng trung trc ca GD.

0,5

Phn B:Cu 5: (01 im)

Ta c:( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )( )

2 2 2

2 2 2

2

4 4 4

4 4 4

4 4 4

14 4 4

x y z xy yz zxx y y z z x

x y xy y z yz z x zx

x y y z z x

x y xy y z yz z x zx

x y y z z x

x y y z z x

x y y z z x

+ + + + + + +

+ + + = + +

+ + +

+ + + = + +

+ + +

= + +

+ + +

Dox >0, y >0, z > 0 nn (1) >0 (pcm)

0,5

0,5

Cu 6: (01 im)

H pt cho tng ng:42 45 35

35 15 14

x y

x y

+ =

=

Trnh by cc bc gii v kt lun h pt c 1 nghim ( )11 13

; ;21 45

x y

=

0,25

0,75

Cu 7: (01 im)

( ) ( ). .CA CB CI IA CI IB= + +

( ) ( )= .CI IA CI IA+

(do I l trung im ca AB)2 2 2 2

2=

=4

CI IA CI IAAB

CI

=

0,25

0,25

0,25

0,25


[vnmath.com]-dethi hoc ki 1 Toan 10-dap an (2010-2011)

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