$9.236 \log _{2}(x-1)-\log _{2}(x+1)+\log _{\frac{x+1}{x-1}} 2>0$.

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$9.236 \log _{2}(x-1)-\log _{2}(x+1)+\log _{\frac{x+1}{x-1}} 2>0$.
Solution. The inequality is equivalent to the following system:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - 1 > 0 , } \\
{ x + 1 > 0 , } \\
{ \operatorname { log } _ { 2 } \frac { x - 1 } { x + 1 } - \frac { 1 } { \operatorname { log } _ { 2 } \frac { x - 1 } { x + 1 } } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{\left(\log _{2} \frac{x-1}{x+1}-1\right)\left(\log _{2} \frac{x-1}{x+1}+1\right)}{\log _{2} \frac{x-1}{x+1}}>0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > 1 , } \\
{ [ \begin{array} { l } 
{ \operatorname { log } _ { 2 } \frac { x - 1 } { x + 1 } > 1 , } \\
{ - 1 < \operatorname { log } _ { 2 } \frac { x - 1 } { x + 1 } < 0 } 
\end{array} ] }
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > 1 , } \\
{ [ \begin{array} { l } 
{ \frac { x - 1 } { x + 1 } > 2 , } \\
{ \frac { 1 } { 2 } < \frac { x - 1 } { x + 1 } < 1 } 
\end{array} ] }
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > 1 , } \\
{ [ \begin{array} { l } 
{ x - 1 > 2 ( x + 1 ) , } \\
{ \frac { 1 } { 2 } < \frac { x - 1 } { x + 1 } < 1 } 
\end{array} ] }
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > 1 , } \\
{ [ \begin{array} { l } 
{ x - 1 > 2 x + 2 , } \\
{ \frac { 1 } { 2 } < \frac { x - 1 } { x + 1 } < 1 } 
\end{array} ] }
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > 1 , } \\
{ [ \begin{array} { l } 
{ x > 3 , } \\
{ \frac { 1 } { 2 } < \frac { x - 1 } { x + 1 } < 1 } 
\end{array} ] }
\end{array} \Leftrightarrow x > 3 .\right.
\end{aligned} \Leftrightarrow
$$

Answer: $x \in(3 ;+\infty)$.