\(2\left|3x-1\right|+1=5\)
\(\Rightarrow2\left|3x-1\right|=4\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\)\(\Rightarrow x=1\left(x\in Z\right)\)
Ta có: \(2\left|3x-1\right|+1=5\)
\(\Rightarrow2\left|3x-1\right|=4\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=1\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy \(\left[\begin{matrix}x=1\\x=\frac{-1}{3}\end{matrix}\right..\)
