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55 TTHHII TTHH VV PP NNDi y l 5 thi th i hc ca LAISAC c tp ch Ton Hc v Tui tr ng trong

4 s t S1 n S 4. s 5 hon ton mi, thay th mt b mt file ngun .Nhng cu hi trong cc trn, ban u hon ton mi, nhng thi gian sau ny thy c ri

rc trong nhng quyn sch luyn thi i hc hay thi th ca mt s trng...

S 372.6/2008 S 1

(Thi gian lm bi:180 pht)

PHN CHUNG CHO TT C CC TH SINH.Cu I.(2 im).Cho ng cong c hm s ( )3 22 1y x x m x m= - - - + (1).

1. Kho st s bin thin v v th ca hm s khi m = 1 .2. Trong trng hp hm s (1) ng bin trong tp s thc R, tnh m din tch hnh phnggii hn bi th ca hm s (1) v hai trc Ox,Oy c din tch bng 1 n v din tch.

Cu II.( 2 im).1. Gii phng trnh nghim thc: 1 tan .tan 2 cos 3 .x x x- =

2. Tm tt c cc gi tr ca tham s k phng trnh : xxx kk 2124)1( -=+-+ c nghim.Cu III.( 2 im)

1 .Trong mt phng vi h to Oxy cho elp (E) :x2 + 4y2 = 4. Qua im M(1 2) k hai ngthng ln lt tip xc vi (E) ti A v B. Lp phng trnh ng thng i qua hai im A v B.

2. Cho tam gic ABC tha mn : ( )

5

os2 3 os2 os2 02c A c B c C + + + = . Tnh ba gc ca tam gic.Cu IV.( 2 im).

1. Tnh tch phn : dxexxx

I xsin2

0

2 .cos2

cos2

+=

p

.

2. Cho ba s thc dng x, y, z tha mn iu kin: .1.2 =+ xzxy

Tm gi tr nh nht ca biu thc: .543

z

xy

y

zx

x

yzS ++=

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PHN T CHN:Th sinh chn cu V.a hoc cu V.b.Cu V.a. Theo chng trnh THPT khng phn ban . ( 2 im)

1. Trong khng gian vi h trc to Oxyz cho hai ng thng

(d1) :

=-

=-+

03

042

z

yx (d2):

=-

=+

01

0

x

zy.

Lp phng trnh mt cu c bn knh nh nht tip xc vi c hai ng thng trn.

2. C tt c bao nhiu s t nhin chn c 4 ch s, sao cho trong mi s ch s ng sau lnhn ch s ng lin trc n.Cu 5.b. Theo chng trnh THPT phn ban th im . ( 2 im)

1. Cho hnh chp t gic S.ABCD. y ABCD l hnh vung cnh bng a, SA vung gc vi mtphng(ABCD) v SA = a. Tnh din tch ca thit din to bi hnh chp vi mt phng qua A vunggc vi cnh SC.

2. Gii bt phng trnh : ( ) 2log3log 12 xx - ( )Rx .

Ht

HNG DN GII.Cu I.1.Bn c t gii.

2. Ta c y = 3x2 4x m + 1.

hm s ng bin trong tp s thc R khi1

' 03

y x R m " - (2).

Phng trnh honh giao im ca th (1) vi trc Ox: ( )3 22 1x x m x m- - - + = 0(x 1)(x2x m ) = 0 th (1) lun ct trc honh ti im c nh

(1 0 ). Mt khc v hm s l hm bc ba c h s cao nht a = 1 > 0 , li ng bin trong R nn th lun ct trc tung c tung m.

Hay khi 31

-m ( ) [ ]

3 2

2 1 0 01y x x m x m x= - - - + " .Do , din tch hnh phng gii hn bi th (1) v hai trc ta l:

( )

212

1)1(2

1

0

23 mdxmxmxxS --=+----= .

M S = 16

13-=m (tha iu kin (2)).

Cu II. 1.iu kin :

2

1cos

0cos

02cos

0cos2x

x

x

x

Phng trnh tng ng :cos3x = cos3x.cosx.cos2x.

Hoc :

=

=

=-=

4

3cos

)(0cos0cos3cos403cos 2

3

x

loaxxxx p

pkx +=

6.

Hoc:cosx.cos2x=1 01coscos2 3 =-- xx 0)1cos2cos2)(1(cos 2 =++- xxx

=++

=-

0)1cos2cos2(

0)1(cos

2 xx

xp

pmx

vnxx

mx2

).(01cos2cos2

.2

2=

=++

=

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Vy phng trnh c nghim l : pp

kx +=6

pmx 2= . ),( Zmk .

2. t t = 2x k 10

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MN l ng vung gc chung ca hai ng thng (d1) v (d2) ,ta c

-=

=

=--

=--

=---+

=++--

^

^

1'

1

03'2

06'5

0'3'0

00'46

t

t

tt

tt

ttt

ttt

bMN

aMN.

T suy ra phng trnh mt cu cn tm l :4

9)2()

2

3( 222 =-++- zyx .

2. Gi s s l 4321 aaaax= .Theo yu cu bi ton cc ch s a1, a2, a3, a4 khc nhau tng i mt

v khc khng , v x l s chn nn ta c cc trng hp sau :TH1: a4 = 4 , t yu cu ton s l x = 1234. Do c mt cch chn .TH2: a4 = 6 , t yu cu ton ba s hng a1, , a2 , a3 ch c ly trong tp{ }5,4,3,2,1 v cc ch s

tng dn nn c 35 10C = s cho trng hp ny .

TH3 : a4= 8 ,tng t ba s hng a1, , a2 , a3 cn li ch c ly trong tp{ }7,6,5,4,3,2,1 nn c37 35C = s cho trng hp ny.

Vy c 1+10 + 35 = 46 s c chn theo yu cu ton .Cu Vb.1. Bng phng php ta ,chn A(0,0,0) ,B(a 0 0) D(0 a 0) C(aa 0) S(0 0 a).Gi s mt phng (P) cho ct SB, SC SD ln lt ti E, G , F. Mt phng (P) i qua A v vung

gc SC nn nhn vect )( aaaSC -= lm VTPTphng trnh (P) l :x + y z = 0 .(5)

Ta lp phng trnh ng thng SD

-=

=

=

taz

ty

x 0

(6) . F l giao im ca SD v (P) nn n l nghim

h phng trnh ( 5) v (6) )2

2

0(aa

F . Tng t G l giao im ca (P) v SC )3

2

3

3(

aaaG .

Do din tch thit din AEGF : [ ] .32

)(22a

AFAGAGFdtS ===

2. iu kin : x>1 , 2x .

Ta c ( ) 2log3log 12 xx - xx 22

3 log1

)1(log1

-.

Khi 21 t v tx 2= . Bt phng trnh vit li 14

1

4

3143

+

-

tt

tt (7)

ttt

tf

+

=

4

1

4

3)( l hm s lin tc trong )

2

1( +

Ta c

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S 418. 4/2012

S 2(Thi gian lm bi:180 pht)

PHN CHUNG CHO TT C CC TH SINH( 7, 0 im):Cu I. (2, 0 im).Cho th ( C ) c phng trnh 3 23 2y x x= - + .

1. Kho st v v th ( C ).2. Qua im un I ca th ( C ) vit phng trnh ng thng ( d ) ct th ( C ) ti hai im

A, B khc I sao cho tam gic MAB vung ti M, trong M l im cc i ca th ( C ).Cu II. (2,0 im).

1. Gii phng trnh :22cos 3

tan cot .sin2

xx x

x+ =

2 nh tham s m h phng trnh( )

( )

3 19

3 21

x y x m

y x y m

+ + = -

+ + = +

c nghim.

Cu III.(1, 0 im).Tnh tch phn : I = ++++

1

0

12 2)12( dxexx xx .

Cu IV ( 1,0 im).Cho hnh chp t gic S.ABCD, y ABCD l hnh vung cnh a, mt bn SABl tam gic u v vung gc vi y ABCD. Tnh th tch khi nn c ng trn y ngoi tiptam gic ABC v nh ca khi nn nm trn mt phng (SDC).

Cu V.( 1, 0 im).Tm gi tr nh nht ca biu thc: P=3 3 3

3 3 3

a c b a c b

b a bc c b ac a c ab+ +

+ + +,

trong a, b, c l ba s thc dng ty .PHN RING( 3, 0 im): Th sinh ch c lm mt trong hai phn (phn A hoc B)

A.Theo chng trnh Chun.Cu VI a.(2, 0 im)

1. Trong mt phng, vi h trc ta Oxy lp phng trnh ng trn c bn knhR = 2, c tm I nm trn ng thng ( )

1 : 3 0d x y+ - = v ng trn ct ng thng

( )2 : 3 4 6 0d x y+ - = ti hai im A, B sao cho gc

0120AIB = .

2. Trong khng gian, vi h trc ta Oxyz cho ba im ( ) ( ) ( )

123 , 010 , 10 2 A B C - .

Tm trn mt phng (P): 2 0x y z+ + + =

im M sao cho tng

2 2 2

2 3MA MB MC+ +

c gi tr nhnht.

Cu VII. a ( 1,0 im).Gii phng trnh :os

4tanx=2012c x

p +

.B. Theo chng trnh Nng Cao.Cu VI b.(2, 0 im).

1. Trong mt phng, vi h trc ta Oxy cho hai ng thng ( )1 : 3 3 2 0d x y- - + = v

( )2 : 3 3 2 0d x y+ - - = .

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Lp phng trnh ng thngD ct hai ng thng ( ) ( )1 2,d d ln lt ti B, C sao cho tam gic

ABC u c din tch bng 3 3 (vdt), trong nh A l giao im ca ( ) ( )1 2,d d .

2.Trong khng gian, vi h trc ta Oxyz cho hai ng thng cho nhau

( )11 2 3

:1 2 3

x y zd

- - -= = v ( )2

1:

3 2 1

x y zd

-= = .Lp phng trnh mt phng (P) sao cho khong cch

t ( )1d n (P) gp hai ln khong cch t ( )2d n (P).

Cu VII b ( 1, 0 im).Gii phng trnh:os2x

t anx=2012c

.Ht

HNG DN GIICu I.1.T kho st .

2.Theo cu trn hai im ( )02M , ( )1 0I ln lt im cc i v im un ca th ( C ).

Gi k l h s gc ca ng thng qua I nn c phng trnh ( )

1y k x= - . Phng trnh

honh giao im ca ( C ) v ( d ) l ( ) ( )( )

3 2 23 2 1 1 2 2 0x x k x x x x k- + = - - - - - = .

( d ) ct ( C ) ti hai im A,B khc M th phng trnh 2( ) 2 2 0g x x x k= - - - = (*)

c hai nghim khc 1 0 3(1) 0

gk

gD > > -

.

Gi s hai giao im l ( ) ( )1 1 2 2 , A x y B x y , trong 1 2,x x l hai nghim ca phng trnh (*)

v ( ) ( )1 1 2 21 , 1y k x y k x= - = - .V I l tm i xng ca th ( C ) nn tam gic MAB vung ti M

th ( ) ( )2 2

1 2 1 22 2 5 2 5AB MI x x y y= = - + - =

( ) ( ) ( ) ( )( )

2 22 21 2 1 2 1 21 20 1 4 20k x x k x x x x + - = + + - = .

3 2 1 5 1 53 2 0 2, ,

2 2

k k k k k k - + - -

+ + - = = - = = . So vi iu kin k > -3 ta c ba ng thng

( ) ( ) ( )1 5 1 5

2 1 , 1 12 2

y x y x y x- + - -

= - - = - = -

Cu II.1. K 02sin x . Phng trnh tng ng2 21 os6xos 3 os2x os2x 4cos 2x-5cos2x+1=0

2

cc x c c

+= = 2(cos 2 1)(4 cos 2 4 cos 2 1) 0x x x - + - =

os2x=1c (loi),1 2

cos22

x- -

= (loi),1 2 1 1 2

cos 2 arccos .2 2 2

x x kp - + - +

= = +

2.K 0 , 0x y . Cng v theo v ca hai phng trnh ca h ta c :

( ) ( ) ( )

2

2 3 40 0 3 40 0x y xy x y x y x y+ + + + - = + + + - = .

Gii phng trnh bc hai ny ta c 8x y+ = - (loi), 5x y+ =

Th 5y x= - vo phng th nht ca h ta c4

2

mx

-= . tn ti nghim x th 4m .

V4 6

52 2

m my

- += - = . tn ti nghim y th 6m - .

Vy h c nghim th 6 4m- .

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IO

FE

D

CB

A

S

Cu III.Ta c ...)2(.)12(1

0

11

0

121

0

12 222

++++++ ++=++ dxedxexxdxexx xxxxxx

Dng phng php tng phn ta tnh tch phn ++

1

0

12 dxe xx .

t

=

+=

=

= ++++

xv

dxexdu

dxdv

eu xxxx 1122

).12(

Suy ra dxexxxedxe xxxxxx 11

0

21

0

11

0

1 222 )2()( ++++++ +-= .

Do : 31

0

11

0

12 )()12(22

exedxexx xxxx ==++ ++++Cu IV. Gi E, F ln lt trung im AB v CD suy ra ( )

EF SAB^ .

Gi O l tm ca tam gic u ABC. Trong mt phng (SEF) t O dngng thng song song EF ct SF ti I , suy ra I l nh hnh nn

Ta c:2 2 2

EF= .

EF 3 3 3

OI SOOI a

SE

= = =

Bn knh ng trn y2 2 3 3

OS= .3 3 2 3

R SE a a= = =

Vy th tch ca hnh nn l3

2 21 1 1 2 2.3 3 3 3 27

aV R h a a

p

p p= = =

Cu V. Ta c :

.)(

2

3

3

a

c

c

b

b

a

cbab

aca

bcab

ca

+

=+

=+

Tng t:

2

3

3

bcb a

c ac b ac

a b

=

+ +

2

3

3.

cac b

a ba c ab

b c

=

+ +

t X=

b

a Y=

c

b Z=

a

cX,Y,Z >0 v X.Y.Z=1

2 2 2X Y ZP

Y Z Z X X Y = + +

+ + +.

M4

2 ZY

ZY

X ++

++

4

2 ZX

ZX

Y ++

++ .

4

2

ZYXYX

YX

Z++

++

+

P=2

3

2

222

++

+

++

++

ZYX

YX

Z

XZ

Y

ZY

X.Vy Max(P)=

2

3khi a = b = c.

Cu VI a.1.Ta c tm ( ) ( )1 3I x x d- + .Gi H l hnh chiu ca I xung ng thng ( )2d suy ra

tam gic HIA l na tam gic u c cnh IH = 1.Do ta c

2

3 4( 3) 6( , ) 1 6 5 1 11

5

x xd I d IH x x x

+ - + -= = - + = = =

Vy c hai ng trn ( ) ( )2 2

1 2 4x y- + - = , ( ) ( )2 2

11 8 4x y- + + =

2.Gi ( )

x y z l ta im I sao cho :2 2 1

2 3 0 3 3 2

IA IB IC I

+ + = -

uur uur uur r

Ta c ( ) ( ) ( )2 2 2

2 2 2 2 2 2 22 3 2 3 6 2 3MA MB MC MI IA MI IB MI IC MI IA IB IC+ + = + + + + + = + + +uuur uur uuur uur uuur uur

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2 2 22 3MA MB MC+ + c gi tr nh nht khi v ch khi MI t gi tr nh nht, lc M l hnhchiu ca I xung mt phng (P).

ng thng IM vung gc vi (P) nn c phng trnh tham s2 2 1

, ,3 3 2

x t y t z t= + = + = - +

Giao im M l nghim ca h to bi phng trnh (P) v ng thng (d).

Gii h ta c5 5 13

18 18 9

M - - -

.

Cu VIIa.K: cosx 0, t anx>0 sinx 0 . Phng trnh tng ng1

osx 1 12 sinx osx2 2

1sinx

2

sinx 2012sinx.2012 osx.2012

osx2012

cc

cc

= =

V hm s tanx c chu k kp , tanx>0 ta ch xt min nghim sao cho s inx>0, cosx>0 t suy ramin nghim s inx0,cosx>0 )

tanx>0 vi s inx

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2 2 2 2cos sin sin cossinx =2012 s inx.2012 osx.2012osx

x x x xcc

- =

V hm s tanx c chu k kp , tanx>0 ta ch xt min nghim sao cho s inx>0, cosx>0 t suy ramin nghim s inx

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2. Trong khng gian vi h ta Descartes Oxyz , cho tam gic ABCc ( ) ( ) ( )

111 , 0 1 1 , 3 5 3A B C- - - .

Lp phng trnh ng phn gic trong gc A ca tam gic ABC.

Cu VII.a (1 im)Cho cc s phcztha mn iu kin: 1 2 5z i- + = . Tm s phc w c mun

ln nht, bit rng: w = z+1+i .B. Theo chng trnh Nng Cao

Cu VI.b (2 im)1. Trong mt phng vi h ta Descartes Oxy , cho hai im ( ) ( )

A 34 , 5 3B . Xc nh im M

trn ng Elip ( )2 2

: 18 2

x yH + = sao cho din tch tam gic MAB c gi tr nh nht.

2. Trong khng gian vi h ta Descartes Oxyz, cho hai ng thng ( )11 1 1

:1 2 2

x y zd

- - -= = v

( )21 3

:1 2 2

x y yd

+ -= =

-ct nhau nm trong mt phng ( )P . Lp phng trnh ng phn gic ca gc

nhn to bi ( )1d , ( )2d nm trong mt phng ( )P .

Cu VII.b (1 im) Gii h phng trnh

2

1 20121

2 4 .log 0

x y

y

xy

x

-+ = + + =

, vi n thc 0, 0x y> > .

..Ht

HNG DN GIICu I.a) T gii.

b) tip tuyn ti ( ) ( )0 0 mM x y C song song vi ng thng 3 1y x= - + th h s gc ca tip

tuyn

( )

0' 3y x = -

( )

20

2

31

m

x

- - = -

-

2

0 03 6 1 0x x m - + - =

, 2m -

(1)

Gi thit: 0 03 1

( ) 1010

x yd M d

+ -= =

20 0

20 0

3 12 11 0

3 8 9 0

x x m

x x m

- + + =

+ - + =, (2)

Gii (1) v (2) ta c 1m = hoc43

3m = .

Cu I a)iu kinosx 0

cos4x 0

c

. Phng trnh tng ng ( ) ( )8 2 s inxcos2x= tanx+tan4x + tanxtan4x-1

sin5x cos5x8 2 s inxcos2x=cosxcos4x cosxcos4x

- sin 8 sin 54

x x p = -

2.

12 3 ( ). (*)5 2

.52 13

x k

k Zx k

p p

p p

= - +

= +

So vi iu kin (*) chnh l nghim ca phng trnhb) Cng hai v ca hai phng trnh ca h, ta c:

( ) ( ) ( ) ( )

22 1 3 1 2 1 10 0 1 3 1 10 0x y x y x y x y x y+ + + + + + + - = + + + + + - =

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Gii phng trnh bc hai ny ta c1 5

1 2

x y

x y

+ + = - + + =

thay1 5

1 2

y x

y x

+ = - - + = -

vo mt trong hai phng trnh ca h v tip tc gii ta c nghim ca h

cho l ( ) ( ) ( )

: 0 3 , 1 0x y

Cu IIITch phn vit li ( )2

1

ln 1 ln ln.

e x xI dx

x

+ +=

t lndx

t x dtx

= = khi 1 0 1x t x e t= = = =

Do :

( ) ( )1 11

2 2

200 0

ln 1 ln 11

tI t t dt t t t dt

t = + + = + + - +

1 2

20

1 (1 )ln(1 2) ln(1 2) 1 2

2 1

d t

t

+= + - = + + -

+ .

Cu IV. Dng ( )SH ABCD^ , vSA SB SD= = HA HB HD= = H l tm ng trn ngoi tip tamgic ABD, m tam gic ABD u H va l trng tm, trc tmtam gic ABD.Do HD AB^ , m //AB CD HD DC ^ (1), theo nh l bang vung gc ta cng c SD DC ^ (2). T (1) v (2) gc nhn 030SDH= chnh l gc ca hai mt phng ( )SDC v ( )ABCD .Xt tam gic SHD vung ti H c

0 1 3 3, 30 2 2 2

SD a SDH SH a HD a AB a= = = = = .

Vy th tch hnh chp SABCD l 31 3 3. .3 16ABCD

aV S SH = =

Cu V.Cch 1.t1 1 1 1 1 1

1a b cx y z x y z

= = = + + = v 0, 0, 0x y z> > > .

Biu thc cho vit li2 4 3 9 6 36

x y zP

x y z= + +

+ + +

Ta c1 1 1

1 .2 4 2 3 9 3 6 36 6

x y zP

x y z

- = - + - + -

+ + +

1 1 11

2 3 6P

x y z

- = - + +

+ + + .

Ta li c:1 1 4

2 2x x

+

+

ng thc xy ra khi 2x =

Tng t v cng cc bt ng thc li suy ra ,ta c:1 1 1 1

2 3 4 2x y z

- + + -

+ + +

Do :1 1 1 1 1

12 3 6 2 2

P Px y z

- = - + + -

+ + +

Vy gi tr nh nht1

2P= khi v ch khi

1 1 1, ,

2 3 6a b c= = =

H

D

CB

A

S

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Cch 2. Chn im ri:1 2 4 1 1 3 9 1 1 6 36 1

, , .2 4 16 2 3 9 36 3 6 36 144 6

a b c

a b c

+ + ++ + +

+ + +

Suy ra2 4 3 9 6 36 1

116 36 144 2

a b cP P

+ + ++ + +

Cu VI a1) Phng trnh ng trn vit li2 2

1 21

2 2

x y- + + =

.

t1

sin 2 sin 12

xxa a

-

= = + 2

os = 2 os 22

yc y ca a

+

= - , trong [ ]

0 2 .a p

Do ta im ( ) ( )2 sin 1 2 os 2M c Ca a+ - .Phng trnh ng thng : 8 0AB x y+ - = .

Ta c ( )

2 cos 94

2M ABd

pa

- -

= . ax cos - 1 0 3

4 4MABS M x y

p p

a a p

= - = + = = -

.

Vy im ( ) ( )

0 3M C- th din tch tam gic MAB c gi tr ln nht.

2) Ta c ( )1 2 2 3AB AB= - - - =uuur

, ( )2 4 4 6AC AC= - =uuur

Ly im D l trung im AC ( )

23 1D AC AB AD - =

Nn phn gic trong gc A ca tam gic ABC cng l ng trung tuyn ca tam gic cn ABD

Gi H trung im ( )1 11 1BD H - , do ng phn gic cn tm c phng trnh

1

: 1

1 2

x

AH y

z t

=

= = -

.

Cu VIIa. Xt s phcz x yi= + . T gi thit suy ra ( ) ( )2 2

1 2 5x y- + + = . Suy ra tp hp im

( )M x y biu din s phc z l ng trn tm ( )1 2I - , bn knh 5R = . D dng c c

( )5 sin 1 5 cos 2M a a+ - vi [ ]0 2 .a p Mt khc ( ) ( )w = z+1+i= x+1 1y i+ +

( ) ( ) ( ) ( ) ( )2 22 22

w x+1 1 5 sin 2 5 os 1 10 2 5 2sin osy c ca a a a = + + = + + - = + -

t 2 sin cost a a= - , tn tiath ( ) ( )2 2 21 2 5 5 2sin cos 5t t a a+ - -

Do Max

w 20= khi v ch khi2 1

sin , os = 3 35 5

c x ya a= - = = - . Vy s phc l w 4 2i= -

Cu VIb. 1) t sin 2 2.sin2 2

xxa a= = os = 2 os ,

2

yc y ca a = , trong [ ]0 2 .a p

Do ta im ( ) ( )2 2 sin 2 osM c Ha a .Phng trnh ng thng : 2 11 0AB x y+ - = .

Ta c: ( )

4 os 11

45

M AB

c

d

pa

- -

= .. in cos 1 2 14 4MAB

S M x yp pa a - = = = =

Vy im ( ) ( )

21M H th din tch tam gic MAB c gi tr nh nht

2) D dng nhn thy hai ng thng ( )1d , ( )2d ct nhau ti giao im ( )111I .

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Chn trn ( )1d im ( )233 3M IM = .Phng trnh tham s ( )2d

( ) ( )21 2 1 2 3 23 2

x t

y t N t t t d

z t

=

= - + - + - = -

.

( ) ( )2 1 20

3 9 0 13 , 23 12

tIN IM IN N N

t

== = = - - =

.

Ta c: ( ) ( ) ( )

1 2122 , 1 2 2 , 1 2 2IM IN IN= = - - = -uuur uuur uuur .M

02 2. 1 0 90IM IN MIN= > < uuur uuur gc nhn ca

( )1d , ( )2d chnh l gc 2MIN .Gi ( )231K trung im ca 2MN nn ng thng qua hai im I, K

l ng phn gic ca gc nhn to bi ( )1d , ( )2d c phng trnh

1

: 1 2

1

x t

IK y t

z

= +

= + =

.

Cu VII b, V : x > 0 , y > 0 T (1)

2012 2012 2012

1log log (1 ) log (1 ) (*)

1

xx y y y x x

y

+= - - + = - +

+

t ( ) ( )2012log 1f t t t= - + , xc nh ( )0t" + Ta c :

( )

1'( ) 1 01 ln 2012

f tt

= - >+

vi ( )0t" +

f(t) lun lun ng bin trong ( )0+

Do phng trnh (*) ( ) ( ) ( ) ( )2012 2012log 1 log 1x x y y f x f y x y- + = - + = =

T phng trnh (2) : 2 2 14

12 4 .log 0 4 .log 2 log

4

yy yx x x

+ = = - =

(3)

Th x = y vo phng trnh (3) ta gii phng trnh 14

1log

4

x

x

=

bng phng php v hai th

14

logz x= v 14

xz =

trn cng mt h trc ta vung gc xOz ta thy phng trnh c nghim

duy nht1

2x = . Th li th h phng trnh c nghim

1 1,

2 2x y= =

..

S 440 2/2014

S 4(Thi gian lm bi:180 pht)PHN CHUNGCu 1( 2 im). Cho ng cong ( )mC c phng trnh ( )

3 3 2 5,y mx m x m= - - + - ( m l tham s)

1) Kho st v v th hm s trn khi m = 1.2) Chng minh rng vi mi tham s m ng cong ( )

mC lun lun ct mt ng thng c nh

ti ba im c nh.Cu 2( 1 im).Gii phng trnh 4sin .sin 2 . os3 t anx. tan 2 . os6x x c x x c x= .

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Cu 3 ( 1 im).Gii h phng trnh4 3 1 0

4 (1 )(1 ) 6 1 1 0

x y y

x y x

- + + =

+ + - + + =.

Cu 4( 1 im). Tnh tch phn sau ( )4

2

0

ln(cos )xI e ta nx x dx

p

= + .

Cu 5( 1 im). Cho t din ABCD c , ,AB CD a AC BD b AD BC c= = = = = = v 2 2 2 3a b c+ + = (

n v chiu di ). Tnh th tch nh nht ca t din ABCD .Cu 6( 1 im). Cho a, b, c l ba cnh ca mt tam gic tha mn iu kin 2c b abc+ = .

Tm gi tr nh nht ca biu thc3 4 5

Sb c a a c b a b c

= + ++ - + - + -

.

PHN RING( Th sinh ch c chn mt trong hai phn A hoc B)A. Theo chng trnh Chun

Cu 7a.( 1 im). Trong mt phng, vi h trc ta Oxy cho tam gic ABC c nh (12)A , trng

tm (11)G v trc tm2 10

3 3

H

. Hy xc nh ta hai nh B v C ca tam gic.

Cu 8a( 1 im). Trong khng gian vi h trc ta Oxyz cho tam gic ABC vi

( ) ( ) ( )

1 0 0 , 3 2 2 , 2 0 3A B C - . Lp phng trnh mt cu ( )

C c bn knh nh nht i qua A v lnlt ct hai cnh AB, AC ca tam gic ABC ti hai im E, F sao cho 3, AF 2AE= = .

Cu 9a( 1 im). Gi E l tp cc s t nhin c ba ch s abc ( )0a sao cho ba s a,b,c khc

nhau theo th t tng dn. Tnh xc sut ly ra trong tp E mt phn t l s chn.B. Theo chng trnh Nng Cao

Cu 7b.( 1 im).Trong mt phng vi h trc ta Oxy, cho hai ng thng

1 2( ) : 1 0, ( ) : 1 0d x y d x y+ - = - + = . Lp phng trnh ng trn ( )C ct 1( )d ti A v 2( )d ti hai

im B, C sao cho tam gic ABC l tam gic u c din tch bng 24 3 n v din tch.

Cu 8b.( 1 im). Trong khng gian vi h trc ta Oxyz cho tam gic ABC c nh (120)A ,trng tm (110)G v trc tm

2 10 0

3 3H

. Hy xc nh ta hai nh B v C ca tam gic.

Cu 9b( 1 im). Gii phng trnh ( ) ( ) 12 3 2 3 2x x

x++ + - = .

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

HNG DN V GII.

Cu 1. 1.Khi m = 1 hm s vit li . 3 3 4,y x x= + - T kho st v v

2. Hm s vit li ( )

3

3 1 6 5y m x x x= - + + - .

im c nh ( nu c) m ng cong i qua tha nghim h:3 3 1 0

6 5

x x

y x

- + =

= -.

t 3 2( ) 3 1 0 '( ) 3 3f x x x f x x= - + = = - , khi '( ) 0 1, 1f x x x= = = -V th ca hm s f(x) c ax . ( 1). (1) 3.( 1) 0m minf f f f= - = -

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Cu 2. K cos 0.cos 2 0x x 4sin . sin 2 . os3 tanx. tan 2 . os6 4sin . sin 2 . os2 . os3 tan x.sin 2 . os6x x c x x c x x x c x c x x c x= = . Xy ra:

+ sin 2 02

kx x

p= = , so vi iu kin phng trnh c nghimx mp= .(1)

+ 4 sin . os2 . os3 t anx. os6 sin 4 . os3 s inx. os6 sin x=-sin5x ,3 4 2

k kx c x c x c x x c x c x x x

p p p= = = = - - (2)

T (1) v (2) so vi iu kin phng trnh c nghim ( )3

kx k Z

p= .

Cu 3. K 1, 1x y - - . t 2 1 , 1 0, 0a x b y a b= + = + th h tr thnh2 2 3 3 0

2 3 1 0

a b b

ab a

- + - =

- + =T h phng trnh trn ta bin i v dng s phc sau

( ) ( ) ( ) ( )22 2 23 3 2 3 1 0 3 3 0 3 3 0 ( )a b b ab a i a bi i a bi i z iz i a bi z C- + - + - + = + - + + - = - + - = + =

Gii phng trnh ny, ta c 1 21 , 1 2z i z i= + = - +

Suy ra ( )1 1a b= = do nghim ca h phng trnh trn l3

04

x y

= - =

Cu 4.V4 42

0 0

ln(cos )x xI e ta nxdx e x dx

p p

= + .

Ta tnh4 4 4 4 4

2 42 2

0 0 0 0 0

11 (t anx) 1

o sx o sx

xx x x xe dxJ e ta nxdx e dx e dx e d e

c c

p p p p p

p

= = - = - = + -

M4 4 4

44

00 0 0

(t anx) t anx t anx t anxx x x xe d e e dx e e dx

p p p

pp

= - = - . Nn4

0

1 t anxxJ e dx

p

= -

Ta li c

( )

4 4 4 444

00 0 0 0t anx ln( osx) ln(cos ) ln(cos ) ln 2 ln(cos )

x x x x x

K e dx e d c e x e x dx e e x dx

p p p p

pp

= = - = - + = + .

Do 4

4

0

1 ln 2 ln(cos )xJ e e x dx

p

p

= - - . Vy tch phn 41 ln 2I ep

= - .

Cu 5.

Qua cc nh ca tam gic BCD ta k cc ng thng song song vicc cnh i din, chng ln lt ct nhau tng i mt to thnh tamgic BCD c din tch gp 4 ln din tch tam gic BCD.Ta c CD = 2CD = 2AB suy ra tam gic CAD vung ti A, tng

t ta cng chng minh c hai tam gic CAB, BAD ln ltvung ti A.t ' , ' , 'AB x AC y AD z= = = th ta c

( )

2 2 2

2 2 2 2 2 2 2 2 2

2 2 2

4

4 2

4

x y c

y z a x y z a b c

z x b

+ =

+ = + + = + + + =

D'

C'

B'

D

C

B

A

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( )

( )

( )

2 2 2 2

2 2 2 2

2 2 2 2

2

2

2

x b c a

y a c b

z a b c

= + -

= + -

= + -

( )( )( )2 2 2 2 2 2 2 2 2' ' '1 1 2

4 24 12ABCD AB C DV V xyz b c a a c b a b c= = = + - + - + - .

m ( )( )( )32 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 27 13 27

b c a a c b a b cb c a a c b a b c

+ - + + - + + -+ - + - + - = =

( C-si)

212ABCD

V .Vy th tch ca t din ln nht 212ABCD

V = khi v ch khi 1a b c= = = .

Cu 6. V a, b, c l ba cnh tam gic nn 0, 0, 0a b c a c b b c a+ - > + - > + - > v p dng BT

( )

1 1 4, 0x y

x y x y+ >

+

Ta c1 1 1 1 1 1

2 3Sb c a a c b b c a a b c a c b a b c

= + + + + +

+ - + - + - + - + - + -

2 4 6

c b a + +

Mc khc t gi thit2 1

2c b abc a

b c

+ = + =

Nn ta c2 4 6 1 2 3 3

2 2 4 3ac b a c b a a

+ + = + + = +

.Vy

3 4 54 3S

b c a a c b a b c= + +

+ - + - + -.

Do gi tr nh nht ca 4 3S= , khi v ch khi 3a b c= = = .T CHNA. Theo chng trnh chun

Cu 7a.Gi ( )M x y l ta trung im cnh BC, ta c3 1

12 2

AM AG M

=

uuuur uuur

ng thng BC qua1

12

M

v c VTPT1 4

3 3

AH -

uuurnn c phng trnh BC: 4 1 0x y- + = (1)

Gi ( )O x y l ta tm ng trn ngoi tip tam gic ABC, ta c7 1

3 6 6

OH OG O

= -

uuur uuur

Suy ra ng trn ngoi tip tam gic ABC c tm7 1

6 6

O

-

, bn knh170

36R OA= = c phng

trnh l ( )2 2

7 1 170:

6 6 36C x y

- + + =

(2)

Vy ta im B, C l giao im ca ng thng BC v ng trn ( )C chnh l nghim ca h

hai phng trnh (1) v (2) . Gii h ny ta c ( ) ( )1 0 , 3 :1B C-

Cu 8a. Ta c ( ) ( )

222 , 3,0,3AB AC -uuur uuur m . 0AB AC= uuur uuur tam gic ABC vung ti APhng trnh tham s ng thng AB l 1 , ,x t y t z t= + = = suy ra ta im

( )

2 2 21 3 1, 1E t t t AE t t t t t+ = + + = = = -

Khi ( )1 (211) 111t E AE = uuur

cng hng vi vec t ( )

222ABuuur

v 3 2 2AE AB= < = im E

nm trn cnh AB ca tam gic ABCPhng trnh tham s ng thng AC l 1 ', ', 'x t y t z t= - = = suy ra ta im

( )

2 21 '0 ' ' 0 ' 2 ' 1, ' 1F t t AF t t t t- = + + = = = -

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Khi ( )' 1 (001) 101t F AE = -uuur

cng hng vi vec t ( )3,0,3AC -uuur

v

2 3 2AF AC= < = im F nm trn cnh AC ca tam gic ABCDo mt cu i qua ba im A, E, F c AE v AF vung gc nhau th mt cu c bn knh nh nht

EF 6

2 2R = = v tm I l trung im EF

1 11

2 2I

Vy phng trnh mt cu cn tm l

( )2 2

2 1 1 31

2 2 2x y z

- + - + - =

Cu 9a. Gi s ta c mt s abc tha iu kin cho th r rng a, b, c c ly trong tp

{ }123456789X = v ta c mt cch duy nht sp xp theo ba s c th t tng dn, nn ba s

c chn l mt t hp chp ba ca 9 phn t trong tp X, vy s phn t trong tp E l 39E C=

Chn abc l s chn , ta c cc trng hp sau:+ 8c = th a, b c ly t cc s 1,2,3,4,5,6,7 nn ta c s cch chn trong trng hp ny l 27C .

+ 6c = th a, b c ly t cc s 1,2,3,4,5 nn ta c s cch chn trong trng hp ny l 25C .

+ 4c = th a, b c ly t cc s 1,2,3 nn ta c s cch chn trong trng hp ny l 23C .

Gi A l tp cc bin c s chn ca tp E th s cc phn t ca A l2 2 2

7 5 3A C C C= + + .Do xc su cn tm l :

2 2 27 5 3

39

17

42

A C C C

E C

+ += =

B. Theo chng trnh Nng Cao

Cu 7b.Ta nhn thy hai ng thng 1( )d v 2( )d vung gc nhau ti H(01) nn theo yu cu biton th tm I ca ng trn nm trn ng 1( )d v hai im B, C i xng nhau qua giao im H.

V tam gic ABC u 23

24 3 4 64ABC

S BC BC = = = .

Ta c 3 26 2 4 22 3

AH BC R IA AH= = = = = .

a ng thng 1( )d vit di dng tham s ta tm ca ng trn l ( 1 )I t t- , m

2 21 2 2 8 2 8 2, 23

IH HA IH t t t= = = = = = - . Vy ta tm ng trn l (2 1)I - hoc l

( 23)I - . Vy ta c hai ng trn cn tm l ( ) ( ) ( ) ( )

2 2 2 22 1 32, 2 3 32x y x y- + + = + + - = .

Cu 8b. Mt phng cha tam gic ABC i qua ba im (120)A , (110)G ,2 10

03 3

H

nn c

phng trnh z = 0. Gi ( )M x y z l ta trung im cnh BC, ta c

3 1

1 02 2AM AG M

=

uuuur uuur

Mt phng (P) cha ng thng BC qua1

1 02

M

v nhn VTPT1 4

03 3

AH -

uuurc phng trnh

l: 4 1 0x y- + = . Do phng trnh ng thng BC l giao tuyn ca hai mt phng 4 1 0x y- + =v z = 0.(1)Gi ( )O x y z l ta tm ng trn ngoi tip tam gic ABC, ta c:

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7 13 0

6 6OH OG O

= -

uuur uuurmt cu ngoi tip tam gic ABC c tm

7 1 0

6 6O

-

, bn knh

170

36R OA= = c phng trnh l ( )

2 2

27 1 170:6 6 36

C x y z

- + + + =

(2)

Vy ta im B, C l giao im ca ng thng BC (1) v ng trn ( )C (2) gii h hai phng

trnh ny ta c ( ) ( )

100 3 :10B C- .

Cu 9b. Phng trnh vit li2 3 2 3

2 02 2

x x + -

+ - =

t2 3 2 3

( ) 22 2

x x

f x + -

= + -

2 3 2 3 2 3 2 3'( ) ln ln

2 2 2 2

x x

f x + + - -

= +

V 2 22 3 2 3 2 3 2 3

''( ) ln ln 02 2 2 2

x x

f x x R + + - -

= + > "

Nn hm s 2 3 2 3 2 3 2 3'( ) ln ln2 2 2 2

x x

f x

+ + - -= +

ng bin trong ( )

- +

Mt khc hm s '( )f x lin tc trong ( )- + v '( ) '( )x f x x f x - - + +

Nn phng trnh '( ) 0f x = c ng mt nghim 0x v f(x) i du qua 0x nn phng trnh f(x) = 0

c nhiu nht hai nghim .Th li ta nhn thy phng trnh cho c ng hai nghim x = 0, x = 1.

.

( Cn mi)

S 5(Thi gian lm bi:180 pht)

Cu 1 (2,0 im).Cho ng cong ( )mC c hm s3 3y x x m= - + ( m l tham s thc).

a) Kho st s bin thin v v th ca hm s khi 2m = .b) nh tham s m qua im un ca th ( )mC k c mt ng thng ( )d to vi

th ( )mC mt hnh phng (H) v ( )d tip tc chn trn hai trc ta mt tam gic (T) sao cho din

tch ca (H) v (T) bng nhau u bng 2 (vdt) .

Cu 2 (1,0 im). Gii phng trnh ( )( )

2

tan .cot 2 1 s inx 4cos 4sin 5 .x x x x= + + -

Cu 3 (1,0 im). Tnh tch phn( )

( )

3

4

ln 4tan

sin2 .ln 2tanx

xI dx

x

p

p

= .

Cu 4 (1,0 im).

a) Trog trng hp khai trin theo nh thc Newton ca biu thc ( )21n

x+ ta c h s cha 8x bng

210. Tnh tng cc h s ca cc s hng c khai trin t biu thc trn theo trng hp .b) Cho cc s phc z tha mn 1 34z - = v 1 2z mi z m i+ + = + + . nh tham s m tn

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ti hai s phc 1 2,z z ng thi tha mn hai iu kin trn sao cho 1 2z z- l ln nht.

Cu 5 (1,0 im). Trong khng gian vi h trc ta Oxyz, qua hai im ( ) ( )1 11 , 0 10M N- - lp

phng trnh mt phnga ct mt cu ( )

2 2 2( ) 2 ( 1) ( 1) 5S x y z+ + + + - = mt thit din ng trn m

din tch hnh trn sinh bi ng trn c din tch S p= .Cu 6 (1,0 im). Cho hnh chp t gic S.ABCD, y ABCD l hnh vung cnh a, cnh bn

( )SA ABCD^ v SA = a. Qua A dng mt phnga vung gc vi SC sao choa ct SC, SB, SD ln

lt ti G, M, N.Tnh theo a th tch khi nn (H), bit rng ng trn y ca (H) ngoi tip t gic AMGN v nhO ca (H) nm trn y ABCD ca hnh chp S.ABCD.Cu 7 (1,0 im). Trong mt phng vi h trc ta Oxy, hy tnh din tch tam gic ABC bitrng hai im (55)H , ( )54I ln lt l trc tm v tm ng trn ngoi tip tam gic ABC v

8 0x y+ - = l phng trnh ng thng cha cnh BC ca tam gic.

Cu 8 (1,0 im). Gii phng trnh nghim thc( ) 2x ln x 2x 2 x 1- + = + .Cu 9 (1,0 im). Cho ba s dng x, y, z tha mn 0 x y z< < - ,

lc 3 nghim ca phng trnh (1) l 0, 3, 3x x k x k= = - + = + .V I l tm i xng ca ng cong ( )mC nn din tch ca hnh phng (H) l:

( )3

23

0

12 3 3

2

k

S kx m x x m dx k +

= + - + - = + ( )21

2 3 2 12

S k k = + = = - (v 3k> - ).

Lc ny ng thng ( )d vit liy x m= - + nn (d) ct hai trc ta ti hai giao im

( ) ( )

0 , 0A m B m . V (T) l tam gic vung cn nn din tch ca (T) l 212

S m=

theo gi thit 2 2, 2S m m= = = - .Vy c hai gi cn tm l 2, 2m m= = - .

Cu 2. iu kin :cos 0

sin 2 0 2

x kx

x

p

.

Ta c ( ) ( )

2 3tan .cot 2 1 s inx 4cos 4sin 5 tan .cot 2 3sin 4sin 1x x x x x x x x= + + - = - -

sin 3 11 tan .cot 2 sin 3 sin 3 sin 3 1 0

cos .sin 2 cos .sin 2

xx x x x x

x x x x

+ = = - =

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Nghim phng trnh xy ra :

hoc sin 3 03

nx x

p= = , so vi iu kin phng trnh c nghim l

2,

3 3x m x m

p p

p p= + = +

hocsin 2 1 sin 2 1

sin 2 . cos 1cos 1 cos 1

x xx x

x x

= = - = "

= = - v nghim

Vy nghim ca phng trnh trn l ( )2

, ,3 3

x m x m m Zp p

p p= + = + .

Cu 3. Ta c:( )

( ) ( )

3 3 3

4 4 4

ln 2 ln 2t anxln 2.

sin 2 .ln 2 t anx sin 2 .ln 2 t anx sin 2

dx dxI dx

x x x

p p p

p p p

+= = +

Tnh( )

( )

( )( )

3 33

44 4

ln 2tanxln 2 ln 2 ln 2 ln 2 3ln 2. . . ln ln(2 tan ) .ln

sin 2 .ln 2 t anx 2 ln 2 t anx 2 2 ln 2

ddxx

x

p p

p

p

p p

= = =

.

Tnh3 3

44

1 1ln(t anx) ln 3

sin 2 2 2

dx

x

pp

pp

= = .

Vyln 2 ln 2 3 1

.ln ln 32 ln 2 2

I

= +

.

Cu 4.a) . Khai trin biu thc trn c s hng th (k+1) l ( )2 ,k knC x k n

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( )( ) ( )

2 2

1 2

1 3463 , 4 3

3 5 3 0

x yM M

x y

- + = - -

- - =.

Vy hai s phc cn tm l 3 46 3 , 4 3z i z i= + = - - .

Cu 5.Mt cu (S) c tm ( 2 11)I - - v bn knh 5R = .Gi r l bn knh ng trn thit din, theo gi thit ta c 2. 1S r rp p p= = = .Gi d l khng cch t I n mt phnga ta c 2 2 2 5 1 2d R r d = - = - = .

Mt phnga qua ( )

0 10N - c dng ( ) ( )

2 2 2Ax 1 0 Ax 0 0B y Cz By Cz B A B C+ + + = + + + = + + .Mt khca qua ( )1 11M - nn tha 0 : Ax 0A C By Az Ba+ = + - + = .

V 2 22 2

3( , ) 2 4 2

2

A Ad d I A B

BA Ba

-= = = = =

+( v 2 2 2 0A B C+ + )

Do c hai mt phnga cn tm l : 2 2 1 0x y z+ - + = , 2 2 1 0x y z- - - = .

Cu 6. Ta c ( )

BC SABC SAB BC AM

BC AB

^ ^ ^

^( v ( )AM SAB ) (1)

Mt khc SC SC AM a^ ^ ( vAM a ) (2)

T (1) v (2) suy ra ( )AM SBC AM MG^ ^ ( v ( )MG SBC )AMG D vung ti M, tng t ta cng c tam gic ANGD vungti N tm H ng trn y ca (H) l trung im AG, c bn

knh2

AGR= . Xt tam gic vung SAC ti A c

. 6 6

3 6

SA ACAG a R a

SC= = = .

V OH l ng cao (H) / /OH OH SC Oa ^ l giao im hai ng cho AC, BD1

2OH CG = . Xt tam gic vung SAC c AG l ng cao , nn

2 2 3

33

ACCG a OH a

SC= = =

Vy th tch hnh nn l( )

2 31 3.3 54H

V R OH ap p= = .

Cu7 Ko di ng cao AH ln lt ct BC v ng trn ngoi tip tam gic ABC ti hai imE v K, ta d dng chng minh c E l trung im HK.ng caoAH BC^ nn c phng trnh 0x y- = , E l giao im ca BC v AH (44)E v H l

trung im HK (33)K , suy ra bn knh ng trn ngoi tip tam gic ABC l 5R IK= =

phng trnh ng trn l ( ) ( )

2 25 4 5, ( )x y C- + - =

Vy hai im B, C l nghim ca h hai phng trnh ng thng BC v ng trn( ) (3 5), (6 2)C B C v nh A l nghim h ca ng cao AH v ng trn ( ) (66)C ADin tch tam gicABCl

( )

6 6 81 1, . .3 2 6

2 2 2ABCS d A BC BC

+ -= = = (vdt).

Cu 8. iu kin 0x > ta c( ) ( )22

x 1x ln x 2x 2 x 1 x ln x

2x 2

+- + = + - =

+

Xt hm s2

x 1f(x)

2x 2

+=

+

/ /

2 2

1 xf (x) f (x) 0 x 1

(x 1) 2x 2

- = = =

+ +

H

N

G

M

O

S

D

CB

A

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Lp bng bin thin ta c ( ) 1, 0f x x " > , ng thc xy ra khi x = 1.

Xt hm s1 1

( ) ln '( ) 1 '( ) 0 1x

g x x x g x g x xx x

-= - = - = = = .

Lp bng bin thin ta c ( ) 1, 0g x x " > , ng thc xy ra khi x = 1.Vy phng trnh c ng mt nghim x = 1.

Cu 9 Ta c

3 3

215

x yy z

zP x y x y zxy z y z x

= + + + + +

. t , , . . 1, 1.x y z

a b c a b c cy z x= = = = >

Biu thc vit li3 3

2 15a bP ca b a b c

= + + ++ +

Ta c ( )

3 33 3 1a ba b ab a b ab

a b a b c+ + + =

+ +( v a, b > 0 ).

Vy ( )2 21 15 16

( ), 1P c c f c cc c c

+ + = + = " +

Ta c2

16'( ) 2 '( ) 0 2f c c f c c

c= - = =

Lp bng bin thin ta c ( ) (2) 12,f c f = khi v ch khi1

2 2 22

c a b z y x= = = = = .

Vy gi tr nh nht 12P= khi v ch khi 2 2z y x= = .

+ chc bn c vui, khe v nht l cc em 12 nm nay 20014-2015 s thnh cng tt p trong ccma thi sp ti.+ cui (S 5), mnh gi cho Ton Hc Tui tr ri, c g sai st mong cc bn gp mnh b sung he he he( Bi vit , thng thng ta son bo phi nhn trc, trc khi ln mng nhng vv hc sinhthn yu he he he )

Nguyn Li