The equation $x^{10}+{\left(11x-1\right)}^{10}=0$ has 10 complex roots   $r_1,{\overline{r}}_1,r_2,{\overline{r}}_2,r_3,{\overline{r}}_3,r_4,{\overline{r}}_4,r_5,{\overline{r}}_5$

The value of $\frac{1}{r_1{\overline{r}}_1}+\frac{1}{r_2{\overline{r}}_2}+\frac{1}{r_3{\overline{r}}_3}+\frac{1}{r_4{\overline{r}}_4}+\frac{1}{r_5{\overline{r}}_5}$ is equal to R then $\frac{R}{122}$ is equal to 

$x^{10}+{\left(11x-1\right)}^{10}=0$

${\left(11-\frac{1}{x}\right)}^{10}=0$

consider $\omega =11-\frac{1}{x}$

$$\begin{gathered} \omega =cis \left(\frac{2k\pi +\pi }{10}\right),k=0,1,2\_ \_ \_ \_ \_\_90 \\ \frac{1}{x}=11-\omega \\ \frac{1}{x\overline{x}}=\left(11-\omega \right)\left(11-\overline{\omega }\right)=121+\omega \overline{\omega }-11\left(\omega +\overline{\omega }\right) \\ =122-11\left(\omega +\overline{\omega }\right) \\ \frac{1}{r_1\overline{r_1}}+\frac{1}{r_2\overline{r_2}}\frac{1}{r_3\overline{r_3}}-------+\frac{1}{r_5\overline{r_5}}=5.122-11 \\ \left({\omega }_1+\overline{{\omega }_1}+{\omega }_2+\overline{{\omega }_2}+{\omega }_3+\overline{{\omega }_3}+{\omega }_4+\overline{{\omega }_4}+{\omega }_5+\overline{{\omega }_5}\right) \\ R=5.122 \\ \frac{R}{122}=5 \end{gathered}$$