akiraqueen

$\boxed{\text{Lời giải}}$

$*GTNN$: $2x+3y\leq 6\Leftrightarrow 2x-6\leq -3y\Leftrightarrow \frac{2}{3}x-2\leq -y$

$\Rightarrow K\geq x^2-2x+\frac{2}{3}x-2=(x-\frac{2}{3})^2-\frac{22}{9}\geq \frac{-22}{9}$

Dấu "=" xảy ra khi: $x=\frac{2}{3}, y=\frac{14}{9}$

$*GTLN$: $2x+y\leq 4\Leftrightarrow 2x^2+xy\leq 4x\Leftrightarrow x^2-2x\leq \frac{-xy}{2}\Rightarrow K\leq \frac{-xy}{2}-y=\frac{-y(x+2)}{2}\leq 0$ (Do $y \geq 0$, $x \geq 0$)

Dấu '=' xảy ra khi: $x=0,y=0$ hoặc $x=2,y=0$