\(B=5x^2+4xy-2\left(x-2y\right)+2y^2+3\)
\(=5x^2+4xy-2x+4y+2y^2+3\)
\(=\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)-2\)
\(=\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2-2\)
Nhận xét :
\(\left\{{}\begin{matrix}\left(2x+y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2-2\ge-2\)
\(\Leftrightarrow B\ge-2\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(B_{min}=-2\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
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