Question

Solve the following inequality:
$\frac{|x-1|}{x+2}<1$

Answer 1

$\frac{|x-1|}{x+2}<1$

$\left(i\right)when(x-1)>0,x>1$

$\frac{x-1}{x+2}<1$

$\frac{x-1-x-2}{x+2}<0$

$\frac{-3}{x+2}<0$

$\frac{1}{x+2}>0$

$x\epsilon (2,\infty )$

$\left(ii\right)when(x-1)<0,x<1$

$\frac{-x+1}{x+2}<0$

$\frac{1-x-x-2}{x+2}<0$

$\frac{-2x-1}{x+2}<0$

$\frac{x+\frac{1}{2}}{x+2}>0,x\epsilon (-\infty ,-2)\bigcup (\frac{-1}{2},\infty )$

$when<1,x\epsilon (-\infty ,-2)\bigcup (\frac{-1}{2},1)$

$x\epsilon (-\infty ,-2)\bigcup (\frac{-1}{2},1)\bigcup (2,\infty )$