[tex]|x-1| (|2x-1|-5)\ \textless \ 0\\ \\ \boxed{1}\quad |x-1| \geq0 ,\quad \forall x\in \mathbb_{R} $ \\ \\ Singura conditie care ne intereseaza la $|x-1|$ este sa fie diferit de 0. \\ \\ $ |x-1| \neq 0 \Rightarrow x-1\neq 0 \Rightarrow x \neq 1 \\ \\ \boxed{2}\quad $Cum $ |x-1| $ $ \ \textgreater \ 0, $ trebuie sa punem conditia pentru $|2x-1|-5, \\ $ sa fie $ \ \textless \ 0. \\ \\ \\ |2x-1|-5\ \textless \ 0 \Big|+5 \Rightarrow |2x-1| \ \textless \ 5 \Rightarrow -5\ \textless \ 2x-1\ \textless \ 5 \Big|+1 \Rightarrow \\ \\ \Rightarrow -4\ \textless \ 2x\ \textless \ 6\Big|:2 \Rightarrow -2\ \textless \ x\ \textless \ 3 \Rightarrow x\in (-2,3)[/tex]
[tex]\\ $Din \boxed{1} \cap $ $\boxed{2} \Rightarrow \boxed{x\in (-2,1)\cup(1,3)}\rightarrow \it solutie~ finala.[/tex]
