Ta có :
\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow M_{min}=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)
Vậy ...
\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-2.x\frac{1}{2}+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10,75\)
\(=\left(x-\frac{1}{2}\right)^2+\left(x+3\right)^2+10,75\ge10,75\)
\(MinM=10,75\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}}\)
\(M=\) \(x^2-2x\frac{1}{2}+\frac{1}{4}+y^2+2y3+9+10\)\(-9-\frac{1}{4}\)\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)\(\ge\frac{3}{4}\)
Dấu "=" xảy ra khi : \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\); \(y+3=0\Rightarrow y=-3\)
Vậy GTNN của \(M=\frac{3}{4}\)khi \(x=\frac{1}{2};y=-3\)