Cho nên: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|x+1\right|=0\\\left|2-x\right|=1\end{matrix}\right.\\\left\{{}\begin{matrix}\left|x+1\right|=1\\\left|2-1\right|=0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=0\\2-x=\pm1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=\pm1\\2-x=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\x=\left\{{}\begin{matrix}1\\3\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{{}\begin{matrix}0\\-2\end{matrix}\right.\\x=2\end{matrix}\right.\end{matrix}\right.\)
