\(P=2x^2+y^2-2xy-4x-2y+10\\ P=x^2+y^2-2xy+2x-4x-2y+1+4+5\\ P=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+1+\left(x^2-4x+4\right)+5\\ P=\left(x-y\right)^2+2\left(x-y\right)+1+\left(x-2\right)^2+5\\ P=\left(x-y+1\right)^2+\left(x-2\right)^2+5\\ Do\text{ }\left(x-y+1\right)^2\ge0\forall x;y\\ \left(x-2\right)^2\ge0\forall x\\ \Rightarrow\left(x-y+1\right)^2+\left(x-2\right)^2\ge0\forall x;y\\ \Rightarrow P=\left(x-y+1\right)^2+\left(x-2\right)^2+5\ge5\forall x;y\\ \text{ Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y+1=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-2+1=0\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ \text{Vậy }P_{\left(Min\right)}=5\text{ }khi\text{ }\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)