9.030. $\frac{1}{x+2}<\frac{3}{x-3}$.
Solution.

We have

$\frac{1}{x+2}-\frac{3}{x-3}>0$.

Using the interval method, we get $x \in\left(-\frac{9}{2} ;-2\right) \cup(3 ; \infty)$.

$\xrightarrow[-\sqrt{+} \sqrt{-2} \sqrt{t}]{3} \sqrt{3}$

Answer: $x \in\left(-\frac{9}{2} ;-2\right) \cup(3 ; \infty)$.