Đặt: \(S=2x^2+5y^2+4xy+8x-4y-100\)
\(=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)
\(=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)
Ta có: \(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\)
=> \(\left(x+2y\right)+\left(x+4\right)^2+\left(y-2\right)^2\ge0\)
=> \(\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge-120\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x+2y=0\\x+4=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Vậy MinS = -120 ⇔\(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
