$9x^2+7y^2-12xy-6x+4y=0$$\Leftrightarrow (3x-2y)^2-2(3x-2y)+3y^2=0$
$\Leftrightarrow \mathbb A^2-2\mathbb A=-3y^2 \le 0$
$\Leftrightarrow 0 \le \mathbb A \le 2.$
Vậy: $\color{red}{\boxed{\min \mathbb A=0}}$ khi $\color{green}{x=y=0.}$
$\color{red}{\boxed{\max \mathbb A=2}}$ khi $\color{green}{x=\frac{2}{3};y=0.}$
$$\color{red}{\boxed{\boxed{\bigstar H_2T \bigstar}}}$$