Xét \(2M=2x^2+4xy+4y^2-2x+10\)
\(=\left(x^2+4xy+4y^2\right)+x^2-2x+1+9\)
\(=\left(x+2y\right)^2+\left(x-1\right)^2+9\ge9\)
\(\Rightarrow M\ge\frac{9}{2}\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)
Vậy..
\(M=\frac{1}{2}\left(x^2+4xy+4y^2\right)+\frac{1}{2}\left(x^2-2x+1\right)+\frac{9}{2}\)
\(M=\frac{1}{2}\left(x+2y\right)^2+\frac{1}{2}\left(x-1\right)^2+\frac{9}{2}\ge\frac{9}{2}\)
\(M_{min}=\frac{9}{2}\) khi \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)