\(P=2x^2+y^2-10x-2xy+2019\)
\(P=x^2-2xy+y^2+x^2-10x+25+1994\)
\(P=\left(x^2-2xy+y^2\right)+\left(x^2-2\cdot x\cdot5+5^2\right)+1994\)
\(P=\left(x-y\right)^2+\left(x-5\right)^2+1994\ge1994\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=5\end{cases}\Rightarrow}x=y=5}\)
Vậy.....