$x^2+x-2$
$=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}$
$=\Big(x+\dfrac{1}{2}\Big)^2-\dfrac{9}{4}$
Ta nhận thấy: $\Big(x+\dfrac{1}{2}\Big)^2\ge 0$
$\to \Big(x+\dfrac{1}{2}\Big)^2-\dfrac{9}{4}\ge -\dfrac{9}{4}$
$\to$ Dấu "=" xảy ra khi $x+\dfrac{1}{2}=0$
$\to x=-\dfrac{1}{2}$
$\to \min{A}=-\dfrac{9}{4}$ khi $x=-\dfrac{1}{2}$