## Solution.

$$
\begin{aligned}
& \log _{5} x+1-\log _{x} 30
\end{aligned}
$$

1) Let $x>1$, then $\log _{3} x>0,2 \log _{5} x+1>0$, and the inequality is equivalent to the following:

$\left\{\begin{array}{l}x>1, \\ 1-\log _{3} x>0\end{array} \Leftrightarrow 10$, and the inequality is equivalent to the following:

$\left\{\begin{array}{l}0<x<1, \\ 2 \log _{5} x+1<0\end{array} \Leftrightarrow\left\{\begin{array}{l}0<x<1, \\ \log _{5} x<\log _{5} 5^{-\frac{1}{2}}\end{array} \Leftrightarrow 0<x<\frac{1}{\sqrt{5}}\right.\right.$.

Answer: $\quad x \in\left(0 ; \frac{\sqrt{5}}{5}\right) \cup(1 ; 3)$.
$9.216 \log _{5} x+\log _{x} \frac{x}{3}<\frac{\log _{5} x\left(2-\log _{3} x\right)}{\log _{3} x}$.