9. (14 points) The equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{1}, \overline{r_{1}}, r_{2}, \overline{r_{2}}, r_{3}, \overline{r_{3}}, r_{4}, \overline{r_{4}}, r_{5}, \overline{r_{5}}$, where $\overline{r_{i}}$ is the complex conjugate of $r_{i}$ $(i=1,2,3,4,5)$, find the value of $\sum_{i=1}^{5} \frac{1}{r_{i} \bar{r}_{i}}$.
9. Solution: From $x^{10}+(13 x-1)^{10}=0$, we get $\left(\frac{x}{13 x-1}\right)^{10}=-1$,

Let the ten complex roots of $y^{10}=1$ be $\omega_{i}, \overline{\omega_{i}}(i=1,2,3,4, 5)$,

Then $\frac{x}{13 x-1}=\omega_{i}$ or $\omega_{i}(i=1,2,3,4,5)$, each of which is a linear equation with one root.
Assume $\frac{r_{i}}{13 r_{i}-1}=\omega_{i}$, then $\frac{\bar{r}}{13 \overline{r_{i}}-1}=\overline{\omega_{i}}$,
Solving, we get $r_{i}=\frac{\omega_{i}}{13 \omega_{i}-1}, \overline{r_{i}}=\frac{\overline{\omega_{i}}}{13 \overline{\omega_{i}}-1}$, thus $\frac{1}{r_{i} \overline{r_{i}}}=$
$$
\begin{array}{l}
\frac{\left(13 \omega_{i}-1\right)\left(13 \overline{\omega_{i}}-1\right)}{\omega \overline{\omega_{i}}}=170-13\left(\omega_{i}+\overline{\omega_{i}}\right) . \\
\text { Therefore, } \sum_{i=1}^{5} \frac{1}{r_{i} \overline{r_{i}}}=5 \times 170-13 \sum_{i=1}^{5}\left(\omega_{i}+\overline{\omega_{i}}\right)=850 .
\end{array}
$$