Example 3 Solve the inequalities:
(1) $\frac{1}{x-1} \leqslant x-1$;
(2) $\frac{1}{4-x}+\frac{2}{2+x} \leqslant 1$.
(1) Since $\frac{1}{x-1} \leqslant x-1 \Leftrightarrow \frac{1}{x-1}-x+1 \leqslant 0 \Leftrightarrow \frac{-x^{2}+2 x}{x-1} \leqslant 0$
$$
\Leftrightarrow \frac{x(x-2)}{x-1} \geqslant 0 \Leftrightarrow \frac{x(x-2)}{x-1}>0 \text { or } \frac{x(x-2)}{x-1}=0
$$
$\Leftrightarrow x(x-1)(x-2)>0$ or $x=0$ or $x=2$.
Using the root marking method on the number line, as shown in Figure 2-12.
Therefore, the solution set of the original inequality is $[0,1) \cup[2,+\infty)$.
Figure 2-12
(2) Since $\frac{1}{4-x}+\frac{2}{2+x} \leqslant 1 \Leftrightarrow \frac{2}{x+2}-\frac{1}{x-4}-1 \leqslant 0$
$$
\begin{array}{c}
\Leftrightarrow \frac{-x^{2}+3 x-2}{(x+2)(x-4)} \leqslant 0 \Leftrightarrow \frac{(x-1)(x-2)}{(x+2)(x-4)} \geqslant 0 \\
\Leftrightarrow(x+2)(x-1)(x-2)(x-4)>0 \text { or } x=1 \text { or } x=2 .
\end{array}
$$

Using the root marking method on the number line, as shown in Figure 2-13.
Therefore, the solution set of the original inequality is $(-\infty,-2) \cup[1$, 2] $\cup(4,+\infty)$.
Figure 2-13
Solution guidance: Do not arbitrarily multiply both sides by the denominator factor when removing the denominator. Usually, after moving terms to make one side zero, use the property of equivalent transformation to remove the denominator.