9.199. $\frac{\log _{2}(\sqrt{4 x+5}-1)}{\log _{2}(\sqrt{4 x+5}+11)}>\frac{1}{2}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\sqrt{4 x+5}-1>0, \\ \sqrt{4 x+5}+11>0, \\ \log _{2}(\sqrt{4 x+5}+11) \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\sqrt{4 x+5}>1, \\ \sqrt{4 x+5}>-11, \\ \sqrt{4 x+5}+11 \neq 1\end{array} \Leftrightarrow\right.\right.$

$$
\Leftrightarrow \sqrt{4 x+5}>1,4 x+5>1, x>-1
$$

From the condition we have

$$
\begin{aligned}
& \log _{\sqrt{4 x+5}+11}(\sqrt{4 x+5}-1)>\frac{1}{2} \Leftrightarrow \\
& \Leftrightarrow \log _{\sqrt{4 x+5}+11}(\sqrt{4 x+5}-1)>\log _{\sqrt{4 x+5}+11}(\sqrt{\sqrt{4 x+5}+11})
\end{aligned}
$$

For $x>-1$ it is obvious that $\sqrt{4 x+5}+11>1$, so we get

$$
\left\{\begin{array} { l } 
{ \sqrt { 4 x + 5 } + 1 1 > 1 } \\
{ \sqrt { 4 x + 5 } - 1 > \sqrt { \sqrt { 4 x + 5 } + 1 1 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>-1 \\
\sqrt{4 x+5}-1>\sqrt{\sqrt{4 x+5}+11}
\end{array}\right.\right.
$$

Since both sides of the last inequality are non-negative, squaring them, we have

$$
\begin{aligned}
& \left\{\begin{array}{l}
x>-1, \\
(\sqrt{4 x+5})^{2}-2 \sqrt{4 x+5}+1>\sqrt{4 x+5}+11
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x>-1 \\
(\sqrt{4 x+5})^{2}-3 \sqrt{4 x+5}-10>0
\end{array}\right.
\end{aligned}
$$

Solving the second inequality of this system as a quadratic in terms of $\sqrt{4 x+5}$, we find $\sqrt{4 x+5}>5$, from which $x>5$.

Answer: $x \in(5 ; \infty)$.