\(M=4x^2+y^2+1+4xy+4x+2y+6x^2-6x+1\)
\(M=\left(2x+y+1\right)^2+6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-\frac{1}{2}=0\\2x+y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=-2\end{matrix}\right.\)