Example 1 Solve the inequalities:
(1) $\frac{x+2}{x-1}>0$;
(2) $\frac{2 x+1}{(x+2)(x-3)}<0$.
(1) From $\frac{x+2}{x-1}>0$, we get $(x+2)(x-1)>0$, so the solution set of the original inequality is $(-\infty, -2) \cup(1,+\infty)$.
(2) Since $\frac{2 x+1}{(x+2)(x-3)}<0$
$$
\begin{array}{l}
\Leftrightarrow(2 x+1)(x+2)(x-3)<0 \\
\Leftrightarrow(x+2)\left(x+\frac{1}{2}\right)(x-3)<0 .
\end{array}
$$

Using the root marking method on the number line, as shown in Figure 2-10.
Figure 2-10

Therefore, the solution set of the original inequality is $(-\infty, -2) \cup\left(-\frac{1}{2}, 3\right)$.