9.181. $\log _{0.3} \log _{6} \frac{x^{2}+x}{x+4}<0$.
## Solution.

The original inequality is equivalent to the inequality $\log _{6} \frac{x^{2}+x}{x+4}>1 \Leftrightarrow$

$$
\Leftrightarrow \frac{x^{2}+x}{x+4}>6 \Leftrightarrow \frac{x^{2}+x}{x+4}-6>0 \Leftrightarrow \frac{x^{2}+x-6 x-24}{x+4}>0 \Leftrightarrow
$$

$$
\Leftrightarrow \frac{x^{2}-5 x-24}{x+4}>0 \Leftrightarrow \frac{(x+3)(x-8)}{x+4}>0,(x+3)(x-8)(x+4)>0
$$

Using the interval method, we get $x \in(-4 ;-3) \cup(8 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0690.jpg?height=82&width=442&top_left_y=274&top_left_x=461)

Answer: $x \in(-4 ;-3) \cup(8 ; \infty)$.