`|5x-1| =1+x` ( ĐK : `x \ge -1` )
`=>` \(\left[ \begin{array}{l}5x-1 =1+x\\5x -1 = -(1+x) =-1 -x\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}5x -x =1+1\\5x+x =-1+1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=0\end{array} \right.\) $\quad ( \text{TM})$
Vậy `x \in {1/2 ;0}`