9.083. $\frac{1}{3^{x}+5}<\frac{1}{3^{x+1}-1}$.
## Solution.

Domain of definition: $x \neq-1$.

From the condition

$$
\frac{1}{3^{x}+5}-\frac{1}{3 \cdot 3^{x}-1}<0
$$

Since $3^{x}+5>0$ and $3 \cdot 3^{x}-1>0$ for $x \in \mathbb{R}$, this inequality is equivalent to the inequality $\frac{2 \cdot 3^{x}-6}{3 \cdot 3^{x}-1}<0,\left(3^{x}-3\right)\left(3^{x}-\frac{1}{3}\right)<0,3^{-1}<3^{x}<3,-1<x<1$.

Answer: $\quad x \in(-1 ; 1)$.