\(A=3-x^2+2x-\left|y-3\right|=-\left(x^2-2x+1\right)+4-\left|y-3\right|=-\left[\left(x-1\right)^2+\left|y-3\right|\right]+4\)
Mà ta luôn có : \(\begin{cases}\left(x-1\right)^2\ge0\\\left|y-3\right|\ge0\end{cases}\) \(\Rightarrow\left(x-1\right)^2+\left|y-3\right|\ge0\)
\(\Rightarrow-\left[\left(x-1\right)^2+\left|y-3\right|\right]\le0\Rightarrow A\le4\)
Vậy Max A = 4 <=> x = 1 , y = 3