Question

$\frac{2x - 5}{ |x - 1| } \leqslant 1$

Answer 1

$\frac{2x-5}{|x-1|} \leq 1 \\ \left \{ {{ \frac{2x-5}{x-1} \leq 1} \atop { \frac{2x-5}{x-1} \geq -1}} \right. \left \{ {{ \frac{x-4}{x-1} \leq 0} \atop { \frac{3x-6}{x-1} \geq 0}} \right. \left \{ {{x\in(1;4]} \atop {x\in(-\infty;1)\cup[2;+\infty)}} \right. \\ \\ x \in (2;4]$

Answer 2

$\frac{2x-5}{|x-1|} \leq 1\\\\ \frac{2x-5}{|x-1|} - 1 \leq 0\\\\ \frac{2x-5-|x-1|}{|x-1|} \leq 0\\\\ \left \{ {{2x-5-|x-1| \leq 0} \atop {x\neq1}} \right. \\\\ \left \{ {{|x-1| \geq 2x-5} \atop {x\neq1}} \right. \\\\ \left \{ {{x-1 \geq 2x-5\ \ \ or\ \ \ x-1 \leq -(2x-5)} \atop {x\neq1}} \right. \\\\ \left \{ {{4 \geq x\ \ \ or\ \ \ 3x \leq 6} \atop {x\neq1}} \right. \\\\ \left \{ {{x \leq 4\ \ \ or\ \ \ x \leq 2} \atop {x\neq1}} \right. \\\\ \left \{ {{x \leq 4} \atop {x\neq1}} \right.$

$x\in(-\infty;\ 1)\cup(1;\ 4]$