5. Solve the inequality $\frac{x+21-7 \sqrt{x+9}}{x^{2}-8 x}>0$ (10 points)

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# Solution:

Domain: $x \in[-9+\infty), \quad x \neq 0, \quad x \neq 8$.

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 $\quad x \in(-\infty ; 0) \cup(0 ; 7) \cup(8 ;+\infty)$.

Considering the domain, we have $x \in[-9 ; 0) \cup(0 ; 7) \cup(8 ;+\infty)$

Answer: $x \in[-9 ; 0) \cup(0 ; 7) \cup(8 ;+\infty)$