\(M=x^2+y^2-x+6y+10\)
\(M=x^2-2.\frac{1}{2}x+\frac{1}{4}+y^2+6y+9+1-\frac{1}{4}\)
\(M=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+1-\frac{1}{4}\)
\(M_{min}=1-\frac{1}{4}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2},y=-3\)
P/s tham khảo nha
\(x^2+y^2-x+6y+10\)
=\(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}+y^2+6y+9+\frac{3}{4}\)
=\(\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)
Có \(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\left(y+3\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
\(y+3=0\Rightarrow y=-3\)
Vậy MinM = \(\frac{3}{4}\)\(\Leftrightarrow\)\(x=\frac{1}{2}\)và \(y=-3\)
\(\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)