M=[tex]x^2+y^2+xy-3x-3y[/tex]
=[tex]x^2+y^2+x(y-3)-3y[/tex]
=[tex][x^2+2.x+\frac{y-3}{2}+\frac{(y-3)^2}{4}]-\frac{(y-3)^2}{4}+y^2-3y[/tex]
=[tex][x+\frac{y-3}{2}]^2[/tex]-[tex]\frac{y^2}{4}+\frac{3y}{2}-\frac{9}{4}+y^2-3y[/tex]
=[tex][x+\frac{y}{2}-\frac{3}{2}]^2+\frac{3}{4}(y^2-2y+1)-3[/tex]
=[tex][x+\frac{y}{2}-\frac{3}{2}]^2-\frac{3}{4}(y-1)^2-3[/tex]\geq 3 vs \forall x,y
\Rightarrow Min M=-3\Leftrightarrow
[tex]\left\{ \begin{array}{|} y-1=0 \\ x+\frac{y}{2}-\frac{3}{2}=0 \end{array} right.[/tex]
\Rightarrow x=y=1
mình làm cũng ko dám chắc nữa.
nếu đúng thì nhớ thanks nha!