2 − IA UUR + 2 IB IC UUR UUR − + 2 ID UUR R = 0 5 3 4X X X X = 26 − ΠTAN TAN TAN=  − + − 5 3Û R15 2( 5 - X ; 6 - - Y ; 7 - - Z ) = 0 −Π4TA TỠM ỦỬỤÙC I(5; -6 ; -7 ) = >1 0X=MI 0,25LUỰC ỦOỰ

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

 

5

3

4

x x x x = 26 − π

tan tan tan

=  − + − 

5 3

Û r

15 2

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

 

−

π

4

Ta tỡm ủửụùc I(5; -6 ; -7 )

 = >

1 0

x

=MI

 

0,25

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

0

2

⇔    +    −  ữ  + − =

− uuur + uuur uuuur − + uuuur

2 MA 2 MB MC 2 MD

x m x m

2 5 5 3

ngaộn nhaỏt ⇔

( )

2

3 6 12 2 0 *

0

0

ủoaùn MI ngaộn nhaỏt khi M hc =

^I

^P

Phửụng trỡnh chớnh taộc cuỷa d qua I vaứ d

Caõu V : ( 1 ủieồm )

x − = y + = + z

− + − + =

2 5 1 0

x m y

 

vuoõng goực vụựi (P) : 5 6 7

Xeựt heọ : ( )

2 3

+ + − =

3 5 4 0



M=(P) I d ị M(9;0;-5)

 − + = − +

5 5 5 15

m x m

⇔ ( )

Caõu VII b ( 1 ủieồm )

− + = −

5 5 5

m y (I)

Nghieọm cuỷa heọ laứ soỏ giao ủieồm cuỷa

TH1 : m ≠ 1

Xeựt haứm soỏ

x m

= -

f t = t + t - + t - t + treõn R

( )

³

^{3 3 ln} (

²

^{2 2} )

MinP = 0 khi 3 vaứ y= 1

m

1 m - 1

-

Ta coự :

TH2 : m = 1

f x t t t R

( ) ⁼ _{− +}

²

^{+ + > ∀ ∈}

ẹaởt : t = -2x – 4y +1

'

2

t t

2 2

2

3 2 0,

Khi ủoự :

Xeựt haứm soỏ g(t) = t treõn R vaứ g

^’

(t)=1 >0, ∀ t

 

∈ R

13

2

15 25 13 15 25 25

= + + =  + ữ + ≥

P t t t

4 2 4 4 13 13 13

 

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

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

MinP = 25

x + y - 13 =

13 khi 2 4 28 0

13 khi t = - 15

⇒ f(y) ≤ f(z) ⇒ g(z) ≤ g(x) ⇒ z ≤ x

KL :

Vaọy : x = y = z = t

t laứ nghieọm cuỷa phửụng trỡnh :

m ≠ 1: MinP = 0 khi 3 vaứ y= 1

3

2 3 ln

2

2 2 0

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

x k R

ỡ = ẻ

ùù ùớ

Haứm soỏ h(t) = ^t

³

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

²

^{- 2 t + 2} )

m=1 : MinP = 25

ù = -

13 khi 7 1

y k

ùùợ

ủoàng bieỏn treõn R (vỡ coự

13 2

h t t t

'

2

2

Caõu VIIa ( 1 ủieồm )

= + +

2

1 3

- + >0,∀ t ∈ R) vaứ

Ta coự : ( ^{a bz cz + +}

²

) ( ^{a bz +}

²

^{+ cz} )

h(1) = 0

= a

²

+ b

²

+ c

²

– ab – bc – ca

(*) coự nghieọm duy nhaỏt t= 1 .

KL: Heọ coự nghieọm duy nhaỏt (1;1;1)

= 1

2 (2a

²

+ 2b

²

+2 c

²

–2 ab – 2bc – 2ca)

= ¹ ₂ ^ _ ( ^{a b −} ) (

²

^{+ − b c} ) (

²

^{+ − c a} )

²

^ _{ ≥} ^0(ẹPCM)

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

t loai

3( )

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

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

(2;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

( 2) 1 ( 2) ( 1)

MA MB a b b

 

do b = 2 khụng thỏa món vậy

 − = − ≠

2 1 , 2

 − = − ≠  −

2 1 , 2 2

a b b b

 − ⇔ 

   − 

b b

2 1

 − + = − + −  + = − + −

2

2

2

2

2

a b b b b

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

  −    ữ 

b

  =

b a

1 2

 − = − ≠  

a b