Bài 2 :

A=x2-4x+1=x2-4x+4-3=(x-2)2-3

Vì (x-2)2≥≥0∀∀x ⇒⇒(x-2)2-3≥≥-3∀x

Vậy min A = -3

B=4x2+4x+11=4(x2+x+11/4)=4(x2+2.x.1/2+1/4+10/4)=4(x+1/2)2+10

=> B min = 10

C=(x-1)(x+3)(x+2)(x+6)

C=(x-1)(x+6)(x+3)(x+2)

C=(x2+5x-6)(x2+5x+6)

Đặt x2+5x+6=t . Ta có:

C= (t-12).t=t2-12t=t2-12+36-36=(t-6)2-36

C= (x2+5x+6-6)2-36=(x2+5x)2-36

Vì (x2+5x)2≥0∀x ⇒⇒(x2+5x)2-36≥-36∀x

Vậy min C= -36

D=5-8x-x2=-(x2+8x-5)=-(x2+8x+16-21)=-[(x+4)2−21][(x+4)2−21]

D=-(x+4)2+21=21-(x+4)2

Vì (x+4)2≥0∀x⇒⇒21-(x+4)2≤21∀x

Vậy max D=21

E=4x-x2+1=-(x2-4x-1)=-(x2-4x+4-5)=-[(x−2)2−5][(x−2)2−5]=-(x-2)2+5=5-(x-2)2

Vì (x-2)2≥0∀x⇒⇒5-(x-2)2≤5∀x

Vậy max E=5

*∀x : với mọi x