$x(3x-4)$
$=3x^2-4x$
$=3(x^2-\dfrac{4}{3}x)$
$=3(x^2-\dfrac{2}{3}.2.x+\dfrac{4}{9}-\dfrac{4}{9})$
$=3(x-\dfrac{2}{3})^2-\dfrac{4}{3}$
Ta thấy: $3(x-\dfrac{2}{3})^2-\dfrac{4}{3}\ge -\dfrac{4}{3}$
$\to$ Dấu "=" xảy ra khi $x-\dfrac{2}{3}=0$
$\to x=\dfrac{2}{3}$
$\to \min{A}=-\dfrac{4}{3}$