Example 4 Solve the inequalities:
(1) $7 x+5+\frac{2}{x-4}>6 x+\frac{2}{x-4}+3$;
(2) $\frac{x^{2}(x-1)(x-2)^{2}}{(x-2)(x-1)^{2}} \leqslant 0$.
(1) Since $7 x+5+\frac{2}{x-4}>6 x+\frac{2}{x-4}+3 \Leftrightarrow 7 x+5>6 x+3$ and $x \neq 4$
$$
\Leftrightarrow x>-2 \text { and } x \neq 4 \text {. }
$$

Therefore, the solution set of the original inequality is $(-2,4) \cup(4,+\infty)$.
(2) Since $\frac{x^{2}(x-1)(x-2)^{2}}{(x-2)(x-1)^{2}} \leqslant 0 \Leftrightarrow \frac{x^{2}(x-2)}{x-1} \leqslant 0$ and $x \neq 2$
$$
\begin{array}{l}
\Leftrightarrow\left(\frac{x^{2}(x-2)}{x-1}<0 \text { or } \frac{x^{2}(x-2)}{x-1}=0\right) \text { and } x \neq 2 \\
\Leftrightarrow\left(x^{2}(x-2)(x-1)<0 \text { and } x \neq 2\right) \text { or } x=0 \\
\Leftrightarrow((x-2)(x-1)<0 \text { and } x \neq 0 \text { and } x \neq 2) \text { or } x=0 \\
\Leftrightarrow(1<x<2 \text { and } x \neq 0 \text { and } x \neq 2) \text { or } x=0 .
\end{array}
$$

Therefore, the solution set of the original inequality is $\{0\} \cup(1,2)$.
Note: Canceling the same fraction on both sides of the inequality, or canceling the same factor in a fraction within the inequality, may change the range of allowable values for the variable.