9.129. $y=\sqrt{\frac{x^{2}-1}{(x+3)(x-4)}-1}+\frac{1}{\log _{8}(x-4)}$.
## Solution.

From the condition, we obtain the system of inequalities

$$
\left\{\begin{array} { l } 
{ \frac { x ^ { 2 } - 1 } { ( x + 3 ) ( x - 4 ) } - 1 \geq 0 , } \\
{ \operatorname { log } _ { 8 } ( x - 4 ) \neq 0 , } \\
{ x - 4 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { x + 1 1 } { ( x + 3 ) ( x - 4 ) } \geq 0 , } \\
{ x - 4 \neq 1 , } \\
{ x > 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+11)(x+3)(x-4) \geq 0 \\
x \neq-3 \\
x \neq 5 \\
x>4
\end{array}\right.\right.\right.
$$

Using the interval method, we have $x \in(4 ; 5) \cup(5 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0659.jpg?height=118&width=441&top_left_y=141&top_left_x=591)

Answer: $x \in(4 ; 5) \cup(5 ; \infty)$.

Solve the inequalities $(9.130-9.205)$: