<=> \(x^2+2x\left(y+1\right)+\left(y+1\right)^2+y^2-6y+9+2004\)
<=>\(\left(x+y+1\right)^2+\left(y-3\right)^2+2004\)
Ta có: \(\hept{\begin{cases}\left(x+y+1\right)^2\ge\\\left(y-3\right)^2\ge0\end{cases}0}\)
=> \(\left(x+y+1\right)^2+\left(y-3\right)^2+2004\ge2004\)
Vậy Max A=2004. Dấu bằng xảy ra <=> \(\hept{\begin{cases}x=-4\\y=3\end{cases}}\)