$A=x^2+5x$
$=(x^2+2.\frac{5}{2}.x+\frac{25}{4})-\frac{25}{4}$
$=(x+\frac{5}{2})^2-\frac{25}{4}$
Vì $(x+\frac{5}{2})^2≥0∀x⇒(x+\frac{5}{2})^2-\frac{25}{4}≥-\frac{25}{4}∀x$
Dấu ''='' xảy ra khi $x+\frac{5}{2}=0⇔x=-\frac{5}{2}$
Vậy $A_{min}=-\frac{25}{4}⇔x=-\frac{5}{2}$.