Ta có : \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(y+x-2\right)^2\ge0\end{matrix}\right.\)
=> \(\left(x+1\right)^2+\left(y+x-2\right)^2\ge0\)
=> \(\left(x+1\right)^2+\left(y+x-2\right)^2+2019\ge2019\)
Vậy MinA = 2019 . Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x+1=0\\y+x-2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)