1. Solve the equation

$$
(x-2020)^{2}+(x-2020)^{10}=2(x-2020)^{12}
$$
Answer: $\{2019 ; 2020 ; 2021\}$.

## Solution:

Let $t=x-2020$, then the original equation can be rewritten as

$$
t^{2}+t^{10}=2 t^{12}
$$

Therefore, $t=0$ or $t= \pm 1$. We will show that there are no other roots:

1) if we assume that $t^{2}>1$, then $t^{10}>t^{2}$ and $t^{12}>t^{2}$.
2) if we assume that $t^{2}<1$, then $t^{10}<t^{2}$ and $t^{12}<t^{2}$.

In both 1) and 2) cases, the equation will not become an identity.

If $t=0$, then $x=2020$; if $t= \pm 1$, then $x=2019$ or $x=2021$.