\(M=x^2+xy+y^2-3x-3\)
\(=\dfrac{1}{4}x^2+xy+y^2+\dfrac{3}{4}x^2-3x-3\)
\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x-1\right)\)
\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x+1-2\right)\)
\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{2}x-1\right)^2-6>=-6\forall x,y\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-1=0\\\dfrac{1}{2}x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}x=-\dfrac{1}{2}\cdot2=-1\end{matrix}\right.\)