-a² + a - 3

= - ( a² - a + 3 )

= - [ a² - 2 . a . $\frac{1}{2}$ + $(\frac{1}{2})^{2}$ + 3 - $(\frac{1}{2})^{2}$ ]

= - [ $(a-\frac{1}{2})^{2}$ + 3 - $\frac{1}{4}$ ]

= - [ $(a-\frac{1}{2})^{2}$ + $\frac{11}{4}$

= - $\frac{11}{4}$ - $(a-\frac{1}{2})^{2}$  ≤ - $\frac{11}{4}$

Dấu " = " xảy ra khi $(a-\frac{1}{2})^{2}$ = 0 ⇔ a - $\frac{1}{2}$ = 0 ⇔ a = $\frac{1}{2}$

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 $x=\dfrac{1}{2}$