\(P=x^2+y^2-4x+2y+5\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\ge0\) với mọi x,y
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy \(Min_P=0\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)