$4 \cdot 50$ Solve the equation
$$
\frac{1}{x^{2}-10 x-29}+\frac{1}{x^{2}-10 x-45}-\frac{2}{x^{2}-10 x-69}=0 \text {. }
$$
(8th American Invitational Mathematics Examination, 1990)
[Solution] Let $x^{2}-10 x-49=t$. Then the original equation becomes
$$
\frac{1}{t+20}+\frac{1}{t+4}-\frac{2}{t-20}=0
$$

Eliminating the denominators, we get
$$
(t+4)(t-20)+(t+20)(t-20)-2(t+20)(t+4)=0,
$$

Expanding and simplifying, we get
$$
-64 t=640,
$$

which is
$$
t=-10 \text {. }
$$

Substituting back, we get
$$
x^{2}-10 x-49=-10 \text {, }
$$

which is
$$
x^{2}-10 x-39=0 \text {, }
$$

Solving, we get
$$
x_{1}=13, x_{2}=-3 \text {. }
$$

Upon verification, $x_{1}, x_{2}$ are both roots of the original equation.