10.1. Solve in $\square$ the equation $2020^{x^{2}-2 x}+\frac{x^{2}-2 x}{2020^{x}}=1$.
Solution: 1) If $x^{2}-2x>0$, then $2020^{x^{2}-2x}>2020^{0}=1$ and $\frac{x^{2}-2x}{2020^{x}}>0$. Therefore, $2020^{x^{2}-2x}+\frac{x^{2}-2x}{2020^{x}}>1$, so the equation has no solutions in this case.
3) If $x^{2}-2x=0$, i.e., $\left[\begin{array}{l}x=0 \\ x=2\end{array}\right.$, then the initial equation is satisfied.

Answer: $S=\{0 ; 2\}$.