# 5. Solution:

$$
\begin{aligned}
& \frac{8\left(x^{3}+27\right) \sqrt{x^{2}+8 x+16}}{\left(x^{2}-3 x+9\right)\left(x^{2}+7 x+12\right)} \geq x+3 \Leftrightarrow \frac{8(x+3)\left(x^{2}-3 x+9\right) \sqrt{(x+4)^{2}}}{\left(x^{2}-3 x+9\right)(x+3)(x+4)} \geq x+3 \Leftrightarrow \\
& x \neq-3, \frac{8 | x+4|}{(x+4)} \geq x+3 \Leftrightarrow\left[\begin{array}{c}
\left\{\begin{array}{c}
x \in(-4 ;-3) \cup(-3 ;+\infty), \\
x \leq 5, \\
x \in(-\infty ;-4), \\
x \leq-11 .
\end{array}\right.
\end{array} \Leftrightarrow x \in(-\infty ;-11] \cup(-4 ;-3) \cup(-3 ; 5]\right.
\end{aligned}
$$
Answer: $x \in (-\infty; -11] \cup (-4; -3) \cup (-3; 5]$.