a/ $4x-x²+3$
$=-(x²-4x-3)$
$=-(x²-4x+4-7)$
$=-(x-2)²+7$
Ta nhận thấy: $-(x-2)²≤0$
$→-(x-2)²+7≤7$
$→$ Dấu "=" xảy ra khi $x-2=0$
$↔x=2$
$→\max A=7↔x=2$
Vậy $\max A=7$ khi $x=2$
b/ $-x²+6x-11$
$=-(x²-6x+11)$
$=-(x²-6x+9+2)$
$=-(x-3)²-2$
Ta nhận thấy: $-(x-3)²≤0$
$→-(x-3)²-2≤-2$
$→$ Dấu "=" xayr ra khi $x-3=0$
$↔x=3$
$→\max B=-2↔x=3$
Vậy $\max B=-2$ khi $x=3$