Đặt: $A=-a^2+a-3$
$A=-(a^2-a+3)$
$↔-(a^2-2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4})$
$↔-(a-\dfrac{1}{2})^2-\dfrac{11}{4}$
Ta nhận thấy: $(a-\dfrac{1}{2})^2≥0$
$→-(a-\dfrac{1}{2})^2≤0$
$→-(a-\dfrac{1}{2})-\dfrac{11}{4}≤-\dfrac{11}{4}$
$→$ Dấu "=" xảy ra khi $a+\dfrac{1}{2}=0$
$→a=\dfrac{1}{2}$
$→A_{max}=-\dfrac{11}{4}$ khi $x=\dfrac{1}{2}$
