Ta có :
\(T=x^2+2xy+2y^2-2x-2y-2\)
\(T=x^2+2xy+y^2+y^2-2x-2y+1-3\)
\(T=\left(x^2+y^2+1+2xy-2y-2x\right)+y^2-3\)
\(T=\left(x+y-1\right)^2+y^2-3\)
Vì \(\left(x+y-1\right)^2\ge0\) với \(\forall x;y\)
\(y^2\ge0\) với \(\forall y\)
\(\Rightarrow\left(x+y-1\right)^2+y^2-3\ge-3\) với \(\forall x;y\)
Dấu "=" xảy ra :\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-1\right)^2=0\\y^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y-1=0\\y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
Vậy \(Min_T=-3\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)