5. Solve the inequality $\quad \frac{|x|+21-7 \sqrt{|x|+9}}{x^{2}-8|x|}>0$
# Solution:

Domain of definition: $x \neq 0, \quad x \neq \pm 8$.

Factorize the numerator:

$$
|x|+21-7 \sqrt{|x|+9}=|x|+9-7 \sqrt{|x|+9}+12=(\sqrt{|x|+9})^{2}-7 \sqrt{|x|+9}+12=(\sqrt{|x|+9}-3)(\sqrt{|x|+9}-4)
$$

We have:

$$
\frac{(\sqrt{|x|+9}-3)(\sqrt{|x|+9}-4)}{|x|(|x|-8)}>0 \Leftrightarrow \frac{(|x|+9-9)(|x|+9-16)}{|x|(|x|-8)}>0 \Leftrightarrow \frac{|x|(|x|-7)}{|x|(|x|-8)}>0
$$

Solving by the interval method, we get $\quad|x| \in(0 ; 7) \cup(8 ;+\infty)$.

We have $x \in(-\infty ;-8) \cup(-7 ; 0) \cup(0 ; 7) \cup(8 ;+\infty)$

Answer: $x \in(-\infty ;-8) \cup(-7 ; 0) \cup(0 ; 7) \cup(8 ;+\infty)$