M= x2+y2-x+6y+10=(y2+6y+9)+(x2-x+1/4)+3/4 = (y+3)2+(x-1/2)2+3/4>= 3/4 khi y=-3;x=1/2
Ta có\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)
\(\Rightarrow M\ge\frac{3}{4}\)\(\forall x;y\)
Dấu = xảy ra khi\(\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}}\)
Vậy MIN \(M=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=-3\)