\(M=x^2+y^2-x+6y+10\)

\(\Leftrightarrow M=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\)

\(\Leftrightarrow M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Do \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\)

Dấu " = " xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-3\end{matrix}\right.\)

Vậy Min A là : \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2};y=-3\)