［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

That is,
$$-64 t=640$$

Substituting, we get
$$t=-10$$

That is
$$\begin{array}{l} x^{2}-10 x-49=-10 \\ x^{2}-10 x-39=0 \end{array}$$

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