34. Suppose that the two roots of the equation
$$
\frac{1}{x^{2}-10 x-29}+\frac{1}{x^{2}-10 x-45}-\frac{2}{x^{2}-10 x-69}=0
$$
are $\alpha$ and $\beta$. Find the value of $\alpha+\beta$.
34. Ans: 10 .
Let $y=x^{2}-10 x-29$. Then $x^{2}-10 x-45=y-16, x^{2}-10 x-69=y-40$. Thus $\frac{1}{y}+\frac{1}{y-16}-\frac{2}{y-40}=0$. Therefore $y=10$. So $y=x^{2}-10 x-29=10$. Hence $x^{2}-10 x-39=$ 0 . Therefore $\alpha+\beta=10$.