4. Solve the equation: $\frac{1}{x^{2}-10 x-29}=$ $+\frac{1}{x^{2}-10 x-45}-\frac{2}{x^{2}-10 x-69}=0$.
4. Let $x^{2}-10 x=y$ and simplify, the original equation becomes $(y-29) \frac{-40}{(y-69)}=\frac{24}{(y-69)(y-45)}$. This equation has a unique solution $y=39$, $\therefore x^{2}-10 x=39$, when $x=13$ it meets the requirement,