Cách giải:
$A=\dfrac{-2a^2-2a+4}{a^2}$
$\to A+\dfrac{9}{4}=\dfrac{-2a^2-2a+4}{a^2}+\dfrac{9}{4}$
$\to A+\dfrac{9}{4}=\dfrac{8a^2-8a+16+9a^2}{4a^2}$
$\to A+\dfrac{9}{4}=\dfrac{a^2-8a+16}{4a^2}$
$\to A+\dfrac{9}{4}=\dfrac{(a-4)^2}{4a^2} \geq 0$
$\to A+\dfrac{9}{4} \geq 0$
$\to A \geq -\dfrac{9}{4}$
Dấu "=" xảy ra khi $a=4$
Vậy $GTNN_A=-\dfrac{9}{4} \leftrightarrow x=4$