5. Solve the inequality $\frac{x+3-3 \sqrt{x+1}}{x^{2}-4 x}>0$ (10 points)
Solution:

Domain of the function: $x \in[-1 ;+\infty), \quad x \neq 0, \quad x \neq 4$.

Factorize the numerator:

$x+3-3 \sqrt{x+1}=x+1-3 \sqrt{x+1}+2=(\sqrt{x+1})^{2}-3 \sqrt{x+1}+2=(\sqrt{x+1}-1)(\sqrt{x+1}-2)$

We have:

$$
\frac{(\sqrt{x+1}-1)(\sqrt{x+1}-2)}{x(x-4)}>0 \Leftrightarrow \frac{(x+1-1)(x+1-4)}{x(x-4)}>0 \Leftrightarrow \frac{x(x-3)}{x(x-4)}>0
$$

Solving using the interval method, we get $\quad x \in(-\infty ; 0) \cup(0 ; 3) \cup(4 ;+\infty)$.

Considering the domain, we have $x \in[-1 ; 0) \cup(0 ; 3) \cup(4 ;+\infty)$

Answer: $x \in[-1 ; 0) \cup(0 ; 3) \cup(4 ;+\infty)$.