# 5. Solution:

$$
\frac{-3\left(x^{2}-4 x+16\right)\left(x^{2}+6 x+8\right)}{\left(x^{3}+64\right) \sqrt{x^{2}+4 x+4}} \leq x^{2}+x-3 \Leftrightarrow \frac{-3\left(x^{2}-4 x+16\right)(x+2)(x+4)}{(x+4)\left(x^{2}-4 x+16\right) \sqrt{(x+2)^{2}}} \leq x^{2}+x-3 \Leftrightarrow
$$

$x \neq-4, \frac{-3(x+2)}{|x+2|} \leq x^{2}+x-3 \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x \in(-2 ;+\infty), \\ x^{2}+x \geq 0,\end{array}\right. \\ \left\{\begin{array}{l}x \in(-\infty ;-4) \cup(-4 ;-2), \\ x^{2}+x-6 \geq 0 .\end{array}\right.\end{array} \Leftrightarrow\right.$

$x \in(-\infty ;-4) \cup(-4 ;-3] \cup(-2 ;-1] \cup[0 ;+\infty)$.
Answer: $x \in(-\infty ;-4) \cup(-4 ;-3] \cup(-2 ;-1] \cup[0 ;+\infty)$.

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