\(\left|x-1\right|\le4\)
\(TH_1:x-1\ge0\Rightarrow x\ge1\\ x-1\le4\\ \Leftrightarrow x\le5\\ TH_2:x-1< 0\Rightarrow x< 1\\ -\left(x-1\right)\le4\\ \Leftrightarrow-x+1\le4\\ \Leftrightarrow-x\le3\\ \Leftrightarrow x\ge-3\)
Vậy \(x\in\left(-3;5\right)\)
\(\Leftrightarrow-4\le x-1\le4\Leftrightarrow-3\le x\le5\)