Ta có: \(P=2x^2+5y^2+4xy+4x-2y+2029\)
\(=x^2+4xy+4y^2+x^2+4x+4+y^2-2y+1+2024\)
\(=\left(x+2y\right)^2+\left(x+2\right)^2+\left(y-1\right)^2+2024>=2024\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+2y=0\\x+2=0\\y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)