Question

Solve $\left|\frac{2x-1}{x-1}\right|>2$

Answer 1

$$\begin{gathered} \mathrm{As},\left|\frac{2x-1}{x-1}\right|>2 \\ \Rightarrow \frac{2x-1}{x-1}<-2\mathrm{or}\frac{2x-1}{x-1}>2\left(\mathrm{As},\left|x\right|>2\Rightarrow x<-2\mathrm{or}x>2\right) \\ \Rightarrow \frac{2x-1}{x-1}+2<0\mathrm{or}\frac{2x-1}{x-1}-2>0 \\ \Rightarrow \frac{2x-1+2x-2}{x-1}<0\mathrm{or}\frac{2x-1-2x+2}{x-1}>0 \\ \Rightarrow \frac{4x-3}{x-1}<0\mathrm{or}\frac{1}{x-1}>0 \\ \Rightarrow \frac{4x-3}{x-1}<0\mathrm{or}x-1>0 \\ \Rightarrow \left[\left(4x-3>0\mathrm{and}x-1<0\right)\mathrm{or}\left(4x-2<0\mathrm{and}x-1>0\right)\right]\mathrm{or}\left[x-1>0\right] \\ \Rightarrow \left[\left(x>\frac{3}{4}\mathrm{and}x<1\right)\mathrm{or}\left(x<\frac{3}{4}\mathrm{and}x>1\right)\right]\mathrm{or}\left[x>1\right] \\ \Rightarrow \left[\left(\frac{3}{4}<x<1\right)\mathrm{or}\varphi \right]\mathrm{or}\left[x>1\right] \\ \Rightarrow \left[\frac{3}{4}<x<1\right]\mathrm{or}\left[x<1\right] \\ \Rightarrow \frac{3}{4}<x<1\mathrm{or}x>1 \\ \therefore x\in \left(\frac{3}{4},1\right)\cup \left(1,\infty \right) \end{gathered}$$