Ta có : \(\left\{{}\begin{matrix}\left|x-2\right|>1\\\left|x-2\right|< 4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>1\\x-2< -1\end{matrix}\right.\\-4< x-2< 4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\\-2< x< 6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2< x< 1\\3< x< 6\end{matrix}\right.\)
Vậy ...
Vì \(x\in Z\Rightarrow\left|x-2\right|\in Z\)
\(1< \left|x-2\right|< 4\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=2\\\left|x-2\right|=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\\x-2=3\\x-2=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=0\\x=5\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{-1;0;4;5\right\}\)
Ta có: \(1< \left|x-2\right|< 4\)
mà \(\left|x-2\right|\in Z\)(vì \(x\in Z\))
nên \(\left|x-2\right|\in\left\{2;3\right\}\)
\(\Leftrightarrow x-2\in\left\{2;-2;3;-3\right\}\)
hay \(x\in\left\{4;0;5;-1\right\}\)(nhận)
Vậy: \(x\in\left\{4;0;5;-1\right\}\)
