\(A=x^2+2y^2-2xy-1\)
\(=\left(x^2-2xy+y^2\right)+y^2-1\)
\(=\left(x-y\right)^2+y^2-1\ge-1\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=y=0\end{matrix}\right.\)
Vậy \(Min_A=-1\Leftrightarrow x=y=0\)