Inequality $\left(\frac{2x-1}{2x+1}\right)^2>0$ The inequality
$$\left(\frac{2x-1}{2x+1}\right)^2>0$$
has the same solution set as 
a) $(2x-1)^2 > 0$
b) $2x-1 \neq 0$
c) $2x-1 > 0$
d) None of the above
Is this equivalent to$$\frac{2x-1}{2x+1}>0 \ ? $$

The answer for your question is shortly "No", because $$\left( \frac { 2x-1 }{ 2x+1 }  \right) ^{ 2 }>0\quad \Rightarrow x\in R-\left\{ \pm \frac { 1 }{ 2 }  \right\} \\ \\ \frac { 2x-1 }{ 2x+1 } >0\quad \Rightarrow x\in \left( -\infty ;-\frac { 1 }{ 2 }  \right) \cup \left( \frac { 1 }{ 2 } ;+\infty  \right) $$
