2 − IA UUR + 2 IB IC UUR UUR − + 2 ID UUR R = 01 0X 02Û R( 5 - X ; 6 - - Y ; 7 - - Z ) = 0⇔    +    −  Ữ  + − =X M X M2 5 5 3( )TA TỠM ỦỬỤÙC I(5; -6 ; -7 )23 6 12 2 0 *0,250 0=MILUỰC ỦOỰ

2)Goùi I thoỷa : 2 − IA uur + 2 IB IC uur uur − + 2 ID uur r = 0

1 0

x

 

0

2

Û r

( ^{5 - x ; 6 - - y ; 7 - - z} ) ^{= 0}

⇔    +    −  ữ  + − =

x m x m

2 5 5 3

( )

Ta tỡm ủửụùc I(5; -6 ; -7 )

2

3 6 12 2 0 *

0,25

0 0

=MI

Luực ủoự : 2 − MA uuur + 2 MB MC uuur uuuur − + 2 MD uuuur

− uuur + uuur uuuur − + uuuur

2 MA 2 MB MC 2 MD

ngaộn nhaỏt ⇔

Caõu V : ( 1 ủieồm )

− + − + =

x m y

2 5 1 0

ủoaùn MI ngaộn nhaỏt khi M hc =

^I

( )

^P

 

Xeựt heọ : ( )

Phửụng trỡnh chớnh taộc cuỷa d qua I vaứ d

3 5 4 0

+ + − =



x − = y + = + z

vuoõng goực vụựi (P) : 5 6 7

 − + = − +

5 5 5 15

m x m

2 3

⇔ ( )

M=(P) I d ị M(9;0;-5)

5 5 5

m y (I)

− + = −

TH1 : m ≠ 1

Caõu VII b ( 1 ủieồm )

x m

= -

MinP = 0 khi 3 vaứ y= 1

Nghieọm cuỷa heọ laứ soỏ giao ủieồm cuỷa

1 m - 1

m

-

Xeựt haứm soỏ

TH2 : m = 1

f t = t + t - + t - t + treõn R

( )

³

^{3 3 ln} (

²

^{2 2} )

ẹaởt : t = -2x – 4y +1

Ta coự :

Khi ủoự :

f x t t t R

( ) ⁼ _{− +}

²

^{+ + > ∀ ∈}

 

' 2

13

2

15 25 13 15 25 25

P t t t

= + + =  + ữ + ≥

2

3 2 0,

2 2

t t

4 2 4 4 13 13 13

 

Xeựt haứm soỏ g(t) = t treõn R vaứ g

^’

(t)=1 >0, ∀ t

MinP = 25

∈ R

x + y - 13 =

13 khi 2 4 28 0

13 khi t = - 15

Haứm f(t) vaứ haứm g(t) cuứng ủoàng bieỏn treõn

KL :

R

m ≠ 1: MinP = 0 khi 3 vaứ y= 1

x ≤ y ⇒ f(x) ≤ f(y) ⇒ g(y) ≤ g(z) ⇒ y ≤ z

⇒ f(y) ≤ f(z) ⇒ g(z) ≤ g(x) ⇒ z ≤ x

ỡ = ẻ

x k R

ùù ùớ

Vaọy : x = y = z = t

m=1 : MinP = 25

y k

ù = -

t laứ nghieọm cuỷa phửụng trỡnh :

13 khi 7 1

ùùợ

13 2

t + t - + t - t + = (*)

3

2 3 ln

2

2 2 0

Caõu VIIa ( 1 ủieồm )

Haứm soỏ h(t) = ^t

³

^{+ 2 t - 3 ln +} ( ^t

²

^{- 2 t + 2} )

Ta coự : ( ^{a bz cz + +}

²

) ( ^{a bz +}

²

^{+ cz} )

ủoàng bieỏn treõn R (vỡ coự

= a

²

+ b

²

+ c

²

– ab – bc – ca

' 2 2

h t t t

= + +

= 1

2

1 3

2 (2a

²

+ 2b

²

+2 c

²

–2 ab – 2bc – 2ca)

- + >0, ∀ t ∈ R) vaứ

h(1) = 0

= ¹ ₂ ^ _ ( ^{a b −} ) (

²

^{+ − b c} ) (

²

^{+ − c a} )

²

^ _{ ≥} ^0(ẹPCM)

(*) coự nghieọm duy nhaỏt t= 1 .

KL: Heọ coự nghieọm duy nhaỏt (1;1;1)

II 2. sin x + sin

²

x + sin

³

x + sin

⁴

x = cos x + cos

²

x + cos

³

x + cos

⁴

x 1,0

TXĐ: D =R

2 3 4 2 3 4

sin x + sin x + sin x + sin x = cos x + cos x + cos x + cos x

− =

⇔ − + + + = ⇔  +   + + = 0,25

x cosx

[ ] ^{sin 0}

(sin ). 2 2(sin ) sin . 0

x cosx x cosx x cosx

2 2(sin ) sin . 0

x cosx x cosx

+ Với sin 0 ( )

0,25

x cosx − = ⇔ = + x π 4 k π k Z ∈

+ Với 2 2(sin + x cosx + ) sin . + x cosx = 0 , đặt t = sin x cosx + (t ∈ −   2; 2 )  

 = −

t

⇔  = − 

được pt : t

²

+ 4t +3 = 0 1

3( )

t loai

0.25

π π

 = + ∈

( )

x k k Z

  = + ∈

 = +

4

⇒  ∈

x m m Z

2 ( )

t = -1

 = − +



= − +

 

.

Cõu

A nằm trờn Ox nờn ^{A a} ( ) ^{;0 ,} B nằm trờn đường thẳng x y − = 0 nờn B b b ( ; ) ,

AVI.1

(2;1)

M ⇒ MA uuur = − − ( a 2; 1), MB uuur = − ( b 2; b − 1)

(1,0 đ)

Tam giỏc ABM vuụng cõn tại M nờn:

uuur uuur

− − − − =

 = 

( 2)( 2) ( 1) 0

a b b

 ⇔ 

. 0

MA MB

 =  − + = − + −

,

 

2 2 2

MA MB a b b

( 2) 1 ( 2) ( 1)

 

do b = 2 khụng thỏa món vậy

 − = − ≠

2 1 , 2

 − = − ≠  −

2 1 , 2 2

a b b b

 − ⇔ 

2 1

   − 

b b

 − + = − + −  + = − + −

2 2 2 2 2

a b b b b

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

  −    ữ 

b

,25

  =

b a

1 2

 − = − ≠  

a b