Lời giải:
$ A = -x^2+2x-10$
$= -(x^2-2x+1) - 9$
$= -(x-1)^2 - 9$
Ta có: $-(x-1)^2≤ 0 ∀ x$
$⇒ -(x-1)^2 -9 ≤ -9$
Dấu "=" xảy khi $-(x-1)^2=0$
$⇔ -(x-1)=0$
$⇔ -x + 1 = 0$
$⇔ -x = -1$
$⇔ x = 1$
Vậy $A_{max} = -9$ khi $x = 1$
$B= 4x^2-4x+10$
$= (4x^2 - 4x + 1) + 1$
$= (2x-1)^2 + 1$
Ta có: $(2x-1)^2 ≥ 0 ∀ x$
$⇒ (2x-1)^2 + 1 ≥ 1$
Dấu "=" xảy khi $(2x-1)^2=0$
$⇔ 2x-1 = 0$
$⇔ 2x = 1$
$⇔ x = \frac{1}{2}$
Vậy $B_{min} = 1$ khi $x = \frac{1}{2}$
$C= -x^2+4x+20$
$= -(x^2 - 4x - 20)$
$= -(x^2 - 4x + 4) + 24$
$= -(x-2)^2 + 24$
Ta có: $-(x-2)^2 ≤ 0 ∀ x$
$⇒ -(x-2)^2 +24 ≤ 24$
Dấu "=" xảy ra khi $-(x-2)^2 = 0$
$⇔ -x + 2 = 0$
$⇔ x = 2$
Vậy $C_{max}= 24$ khi $x = 2$
