$Ta có:A=4a-2a^{2}$
$=4a-2a^{2} -2+2$
$=2-(2a^{2}-4a+2)$
$=2-2(a^{2} -2a+1)$
$=2-2(a-1)^{2} .$
$Vì 2(a-1)^{2} ≥0 ∀x$
$⇔2-2(a-1)^{2} ≤2 ∀x$
$⇔A≤2 ∀x.$
$Dấu\text{"="}\text {xảy}\text {ra}⇔2(a-1)^{2} =0$
$⇔(a-1)^{2}=0$
$⇔a-1=0$
$⇔a=1.$
$Vậy\text{a}=1\text {thì}\text {Max} A=2.$
$Ta\text {có}:B=-2y^{2} -10y+15$
$=-2y^{2}-10y-\frac{25}{2} +27\frac{1}{2}$
$=27\frac{1}{2} -(2y^{2}+10y+\frac{25}{2}$)
$=27\frac{1}{2}-2(y^{2}+5y+\frac{25}{4}$)
$=27\frac{1}{2}-2(y+\frac{5}{2} )^{2}$ .
$Ta\text{ có}:2(y+\frac{5}{2} )^{2} ≥0∀y$
$⇒27\frac{1}{2}-2(y+\frac{5}{2} )^{2}≤27\frac{1}{2}∀y$
$⇒B≤27\frac{1}{2} ∀y.$
$Dấu\text{"="}\text{xảy}\text{ra}⇔2(y+\frac{5}{2} )^{2}=0$
$⇔(y+\frac{5}{2} )^{2}=0$
$⇔y+\frac{5}{2}=0$
$⇔y=-\frac{5}{2} .$
$Vậy\text{ y}=-\frac{5}{2}\text{thì}\text {Max}\text{B}=27\frac{1}{2}.$
