9.149. $\frac{3}{6 x^{2}-x-12}<\frac{25 x-47}{10 x-15}-\frac{3}{3 x+4}$.
Solution.

Rewrite the given inequality as

$$
\begin{aligned}
& \frac{3}{(3 x+4)(2 x-3)}-\frac{25 x-47}{5(2 x-3)}+\frac{3}{3 x+4}0, \frac{75\left(x+\frac{79}{75}\right)(x-2)}{(3 x+4)(2 x-3)}>0 \Leftrightarrow \\
& \Leftrightarrow\left(x+\frac{79}{75}\right)(x-2)(3 x+4)(2 x-3)>0 .
\end{aligned}
$$

Using the interval method, we have \( x \in\left(-\infty ;-\frac{4}{3}\right) \cup\left(-\frac{79}{75} ; \frac{3}{2}\right) \cup(2 ; \infty) \).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0671.jpg?height=172&width=442&top_left_y=1216&top_left_x=397)

Answer: \( x \in\left(-\infty ;-\frac{4}{3}\right) \cup\left(-\frac{79}{75} ; \frac{3}{2}\right) \cup(2 ; \infty) \).