Example 13 The solution to the equation $\frac{1}{x^{2}+x}+\frac{1}{x^{2}+3 x+2}+$ $\frac{1}{x^{2}+5 x+6}+\frac{1}{x^{2}+7 x+12}=\frac{4}{21}$ is $\qquad$ -
(Tenth Zu Chongzhi Cup Junior High School Mathematics Competition)
Solution: After transformation, each fraction in the original equation is in the form of $\frac{1}{n(n+1)}$. If we convert it into partial fractions, the original equation can be simplified.
Let $\frac{1}{n(n-1)}=\frac{A}{n}+\frac{B}{n+1}$,

i.e., $1=A(n+1)+B n$.
Taking $n=0$, we get $A=1$;
Taking $n=-1$, we get $B=-1$.
$$
\therefore \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \text {. }
$$

The original equation can be transformed into
$$
\begin{array}{l}
\frac{1}{x(x+1)}+\frac{1}{(x+1)(x+2)} . \\
+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}=\frac{4}{21} .
\end{array}
$$

Applying (1), we get
$$
\begin{array}{l}
\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2} \\
-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}=\frac{4}{21},
\end{array}
$$

i.e., $\frac{1}{x}-\frac{1}{x+4}=\frac{4}{21}$.
Solving, we get the roots of the original equation $x_{1}=3, x_{2}=-7$.