5. Given the equation $x^{10}+(13 x-1)^{10}=0$ has 10 complex roots $r_{1}, \overline{r_{1}}, r_{2}, \overline{r_{2}}, r_{3}, \overline{r_{3}}$, $r_{4}, \overline{r_{4}}, r_{5}, \overline{r_{5}}$, then $\sum_{r=1}^{5} \frac{1}{\left|r_{i}\right|^{2}}$ is ( ).
A. 750
B. 850
C. 950
D. 1300
5. B

The original equation is $\left(\frac{1}{x}-13\right)^{10}=-1$. For any root $\alpha$, we have $\left|\frac{1}{\alpha}-13\right|=1$, so $5=\sum_{i=1}^{5}\left|\frac{1}{r_{i}}-13\right|^{2}=\sum_{i=1}^{5}\left(\frac{1}{r_{i}}-13\right)\left(\frac{1}{r_{i}}-B\right)=\sum_{i=1}^{5} \frac{1}{\left|r_{i}\right|^{2}}-13 \sum_{i=1}^{5}\left(\frac{1}{r_{i}}+\frac{1}{\bar{r}_{i}}\right)+$ 845. On the other hand, $-1=\left(\frac{1}{r}\right)^{10}-130\left(\frac{1}{r}\right)^{9}+\cdots$, by Vieta's formulas, we get: $\sum_{i=1}^{5}\left(\frac{1}{r_{i}}+\frac{1}{r_{i}}\right)=130$, so the expression $=850$, choose $\mathrm{B}$.