Ta có: \(x^2-2x+y^2-4y+6=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\)
hay \(x^2-2x+y^2-4y+6\ge1\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(min_{\left(x^2-2x+y^2-4x+6\right)}=1\) khi x=1; y=2
