9.169. $\frac{1}{x^{2}-4}+\frac{4}{2 x^{2}+7 x+6} \leq \frac{1}{2 x+3}+\frac{4}{2 x^{3}+3 x^{2}-8 x-12}$.
## Solution.

Rewrite the inequality as

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

$\Leftrightarrow \frac{2 x+3+4 x-8-x^{2}+4-4}{(2 x+3)(x-2)(x+2)} \leq 0, \frac{x^{2}-6 x+5}{(2 x+3)(x-2)(x+2)} \geq 0$,

$\frac{(x-1)(x-5)}{(2 x+3)(x-2)(x+2)} \geq 0 \Leftrightarrow\left\{\begin{array}{l}(x-1)(x-5)(2 x+3)(x-2)(x+2) \geq 0, \\ x \neq-\frac{3}{2}, \\ x \neq \pm 2 .\end{array}\right.$

Using the interval method, we get $x \in\left(-2 ;-\frac{3}{2}\right) \cup[1 ; 2) \cup[5 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0683.jpg?height=126&width=417&top_left_y=1574&top_left_x=389)

Answer: $x \in\left(-2 ;-\frac{3}{2}\right) \cup[1 ; 2) \cup[5 ; \infty)$.