\(A=4x^2+4xy+y^2+x^2-2x+1+y^2+4y+4+2019\)
\(A=\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+2019\ge2019\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2x+y=0\\x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)