\(x^2-2x+y^2+4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\)
Ta thấy: \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
