\(x^2+2\left(y-4\right)x+2y^2-4y=0\)
\(\Delta'=\left(y-4\right)^2-\left(2y^2-4y\right)\ge0\)
\(\Leftrightarrow-y^2-4y+16\ge0\)
\(\Rightarrow-2-2\sqrt{5}\le y\le-2+2\sqrt{5}\)
\(\Rightarrow y_{max}=-2+2\sqrt{5}\)
Khi đó \(x=6-2\sqrt{5}\)
Ta có :
\(A=x^2+2y^2+2xy-8x-4y\)
\(=\left(x^2+2xy+y^2\right)-8\left(x+y\right)+16+\left(y^2-4y+4\right)-20\)
\(=\left(x+y-4\right)^2+\left(y-2\right)^2-20\)
Với mọi x ta có :
\(\left\{{}\begin{matrix}\left(x+y-4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+y-4\right)^2+\left(y-2\right)^2\ge0\)
\(\Leftrightarrow A\ge-20\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=2\)
Vậy..