a/ $A\,=4x^2-12x+10\\\quad =4x^2-12x+9+1\\\quad =(2x-3)^2+1$
Ta có: $(2x-3)^2\ge 0$
$↔A\ge 1$
$→$ Dấu "=" xảy ra khi $2x-3=0$
$↔2x=3\\↔x=\dfrac{3}{2}$
Vậy $A$ đạt GTNN là $1$ khi $x=\dfrac{3}{2}$
b/ $B=2x-x^2-2\\\quad =-x^2+2x-1-1\\\quad =-(x^2-2x+1)-1\\\quad =-(x-1)^2-1$
Ta có: $(x-1)^2\ge 0$
$↔-(x-1)^2\le 0\\↔B\le -1$
$→$ Dấu "=" xảy ra khi $x-1=0$
$↔x=1$
Vậy $B$ đạt GTLN là $-1$ khi $x=1$