Question

Solve $\frac{\left|x+2\right|-x}{x}<2$

Answer 1

$$\begin{gathered} \mathrm{As},\frac{\left|x+2\right|-x}{x}<2 \\ \Rightarrow \frac{\left|x+2\right|-x}{x}-2<0 \\ \Rightarrow \frac{\left|x+2\right|-x-2x}{x}<0 \\ \Rightarrow \frac{\left|x+2\right|-3x}{x}<0 \\ \mathrm{Case}\mathrm{I}:\mathrm{When}x\ge -2,\left|x+2\right|=\left(x+2\right), \\ \frac{\left(x+2\right)-3x}{x}<0 \\ \Rightarrow \frac{2-2x}{x}<0 \\ \Rightarrow \frac{-2\left(x-1\right)}{x}<0 \\ \Rightarrow \frac{x-1}{x}>0 \\ \Rightarrow \left(x-1>0\mathrm{and}x>0\right)\mathrm{or}\left(x-1<0\mathrm{and}x<0\right) \\ \Rightarrow \left(x>1\mathrm{and}x>0\right)\mathrm{or}\left(x<1\mathrm{and}x<0\right) \\ \Rightarrow x>1\mathrm{or}x<0 \\ \Rightarrow x\in [-2,0)\cup \left(1,\infty \right) \\ \mathrm{Case}\mathrm{II}:\mathrm{When}x\le -2,\left|x+2\right|=-\left(x+2\right), \\ \frac{-\left(x+2\right)-3x}{x}<0 \\ \Rightarrow \frac{-x-2-3x}{x}<0 \\ \Rightarrow \frac{-4x-2}{x}<0 \\ \Rightarrow \frac{-2\left(2x+1\right)}{x}<0 \\ \Rightarrow \frac{2x+1}{x}>0 \\ \Rightarrow \left(2x+1>0\mathrm{and}x>0\right)\mathrm{or}\left(2x+1<0\mathrm{and}x<0\right) \\ \Rightarrow \left(x>\frac{-1}{2}\mathrm{and}x>0\right)\mathrm{or}\left(x<\frac{-1}{2}\mathrm{and}x<0\right) \\ \Rightarrow x>0\mathrm{or}x<\frac{-1}{2} \\ \Rightarrow x\in (-\infty ,-2]\cup \left(0,\infty \right) \\ \mathrm{So},\mathrm{from}\mathrm{both}\mathrm{the}\mathrm{cases},\mathrm{we}\mathrm{get} \\ x\in [-2,0)\cup \left(1,\infty \right)\cup (-\infty ,-2]\cup \left(0,\infty \right) \\ \therefore x\in \left(-\infty ,0\right)\cup \left(1,\infty \right) \end{gathered}$$