Ta có: \(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\ge\left|x+2+1-x\right|=3\forall x\)
\(-\left(y+2\right)^2+3\le3\forall y\)
mà \(\left|x+2\right|+\left|x-1\right|=-\left(y+2\right)^2+3\)
nên \(\begin{cases}\left|x+2\right|+\left|x-1\right|=3\\ 3-\left(y+2\right)^2=3\end{cases}\Rightarrow\begin{cases}\left|x+2\right|+\left|1-x\right|=3\\ \left(y+2\right)^2=0\end{cases}\)
=>\(\begin{cases}\left(x+2\right)\left(x-1\right)\le0\\ y+2=0\end{cases}\)
=>\(\begin{cases}-2\le x\le1\\ y=-2\end{cases}\Rightarrow\begin{cases}x\in\left\lbrace-2;-1;0;1\right\rbrace\\ y=-2\end{cases}\)
Vậy: (x;y)∈{(-2;-2);(-1;-2);(0;-2);(1;-2)}
