TH1: \(\left|1-2x\right|=1-2x\)
Ta có: \(2< 1-2x< 5\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-2x>2\\1-2x< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x>1\\-2x< 4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -\dfrac{1}{2}\\x>-2\end{matrix}\right.\Leftrightarrow-2< x< -\dfrac{1}{2}\)
\(\Rightarrow x=-1\)
TH2: \(\left|1-2x\right|=-\left(1-2x\right)=2x-1\)
Ta có: \(2< 2x-1< 5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1>2\\2x-1< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x>3\\2x< 6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< 3\end{matrix}\right.\Leftrightarrow\dfrac{3}{2}< x< 3\)
\(\Rightarrow x=2\)
Vậy \(x\in\left\{-1;2\right\}\)
