2 3X 2 3 6 5X 8 03 − + − − = , ỦIEÀU KIEỌN

2. 2 3x 2 3 6 5x 8 0

³

− + − − = , ủieàu kieọn : 6x x 5−+ vaứ 6 – 5x = 8 5t

³

ẹaởt t =

³

3x 2− ⇔ t

³

= 3x – 2 ⇔ x = t

³

23+ − − =Phửụng trỡnh trụỷ thaứnh : 8 5t

³

2t 3 8 0− = − ⇔

{

^{15tt 4≤}

³

^+4t

²

^{−32t 40 0+ =} ⇔ t = -2. Vaọy x = -2 ⇔ 8 5t

³

3 8 2tCaõu III.

π

π

π

2

2

2

( )

∫ ∫ ∫

3

2

5

2

= − = −I x xdx xdx xdxcos 1 cos cos cos

0

0

0

2

4

2

2

2

2

2

4

( ) ( )

= = − = − +cos cos 1 sin cos 1 2sin sin cosI x xdx x xdx x x xdx

1

= ⇒ =t x dt xdxsin cosπ ⇒ t = 1Đổi cận: x= 0 ⇒ t = 0; x = 2

1

3

5

1

t t2 8

∫

= − + = − + =

2

4

1 2 3 5 15I t t dt t

0

0

π

π

π

π

π

π

π= = + = + = + =

2

2

2

2

2

2

2

I xdx xdx dx xdx x x1 cos 2 1 1 1 1

∫ ∫ ∫ ∫

cos cos 2 sin 2

2

2 2 2 2 4 4

0

0

0

0

π

cos 1 cos 8

3

2

I x xdx15 4

0

Caõu IV. Tửứ giaỷ thieỏt baứi toaựn ta suy ra SI thaỳng goực vụựi maởt phaỳng ABCD, goùi J laứ trung ủieồm cuỷa BC; E laứ hỡnh chieỏu cuỷa I xuoỏng BC.= + = S

CIJ

= ì = = , CJ=BC a 52a a 3aIJ CH 1 3a 3a

2

IJ 2 22 2 2 a 42 = 2

2

2

3a 1 1 3a 3a 6a 3a 3= = ì ⇒ = = ⇒ = = , ⇒ S

CIJ

IE CJ IE SE ,SI4 2 CJ 2 5 5 5 1 1 3a 3 3a 15N

[ ]

³

A B=  + ữ =V a 2a 2a3 2 5 5HI JED C⇔ + + =Caõu V. x(x+y+z) = 3yz 1 y z 3y zx x x x= > = > = + > .Ta cúu v t u vĐặt y 0, z 0, 0x x + u v t+ = ≤  ữ = ⇔ − − ≥ ⇔ − + ≥ ⇔ ≥1 3 3 3 3

2

4 4 0 2 3 2 0 2t uv t t t t t2 4Chia hai vế cho x

³

bất đẳng thức cần chứng minh đưa về

(

^1+u

) (

³

^{+ +1 v}

)

³

^{+3 1}

(

^+u

) (

^{1+v u v}

) (

^{+ ≤}

) (

^{5 u v+}

)

³

( ) ( ) ( ) ( ) ( ) ( ) ( )

3

2

2

3

⇔ + − + + − + + + + + ≤t u v u v u v t t2 3 1 1 3 1 1 3 1 1 5

( ) ( ) ( ) ( )

3

3

3

3

⇔ + − + + ≤ ⇔ + − + + + ≤2 6 1 1 5 2 6(1 ) 5t u v t t u v uv tt t t t t t t t t t2 6 1 1 5 4 6 4 0 2 1 2 0

( ) ( ) ( )

3

3

3

2

⇔ + −  + + ữ≤ ⇔ − − ≥ ⇔ + − ≥ẹuựng do t ≥ 2.PHAÀN RIEÂNG A.Theo chửụng trỡnh ChuaồnCaõu VI.a. 1. I (6; 2); M (1; 5)∆ : x + y – 5 = 0, E ∈ ∆ ⇒ E(m; 5 – m); Goùi N laứ trung ủieồm cuỷa ABx 2x x 12 m = − = − + = −I trung ủieồm NE ⇒

^N

^I

^E

y 2y y 4 5 m m 1 ⇒ N (12 – m; m – 1)

N

I

E

= (m – 6; 5 – m – 2) = (m – 6; 3 – m) = (11 – m; m – 6); IEuurMNuuuuruuuur uur ⇔ (11 – m)(m – 6) + (m – 6)(3 – m) = 0MN.IE 0=⇔ m – 6 = 0 hay 14 – 2m = 0 ⇔ m = 6 hay m = 7+ m = 6 ⇒ MNuuuur = (5; 0) ⇒ pt AB laứ y = 5+ m = 7 ⇒ MNuuuur = (4; 1) ⇒ pt AB laứ x – 1 – 4(y – 5) = 0 ⇒ x – 4y + 19 = 0