$\frac{1}{{{x^2}}} + \frac{1}{{{{(1 - x)}^2}}} = \frac{{160}}{9}$
$ \Rightarrow {\left( {\frac{1}{x} + \frac{1}{{1 - x}}} \right)^2} - 2\frac{1}{{x(1 + x)}} = \frac{{160}}{9}$
$ \Rightarrow {\left( {\frac{1}{{x(1 - x)}}} \right)^2} - 2\left( {\frac{1}{{x(1 - x)}}} \right) = \frac{{160}}{9}$
$\left( {\frac{1}{{x(1 - x)}}} \right) = t$
${t^2} - 2t = \frac{{160}}{9} \Rightarrow {t^2} - 2t + 1 = \frac{{169}}{9}$
${(t - 1)^2} = \frac{{169}}{9} \Rightarrow $
$\left\{ \begin{gathered}
  t - 1 = \frac{{13}}{3} \to t = \frac{{16}}{3} \hfill \\
  t - 1 =  - \frac{{13}}{3} \Rightarrow t = \frac{{ - 10}}{3} \hfill \\ 
\end{gathered}  \right.$
$\frac{1}{{x(1 - x)}} = \frac{{16}}{3} \Rightarrow  - {x^2} + x = \frac{3}{{16}}$
${S_1} = 1$
$\frac{1}{{x(1 - x)}} =  - \frac{{10}}{3} \Rightarrow  - {x^2} + x =  - \frac{3}{{10}}$
${S_1} + {S_2} = 2$