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