Ta có: \(P=4x^2-4xy+5y^2-4x+18y-52\)
\(=4x^2-4xy+y^2-4x+2y+4y^2+16y-52\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)+1+4y^2+16y+16-69\)
\(=\left(2x-y-1\right)^2+4\left(y+2\right)^2-69\ge-69\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}2x-y-1=0\\ y+2=0\end{cases}\Rightarrow\begin{cases}y=-2\\ 2x=y+1=-2+1=-1\end{cases}\Rightarrow\begin{cases}y=-2\\ x=-\frac12\end{cases}\)

