1. Find all solutions to the equation $(x-|x|)^{2}+x+|x|=2020$.
Answer: $\{1010 ;-\sqrt{505}\}$.

## Solution:

1) Let $x \geq 0$, then $(x-x)^{2}+x+x=2020$, we get $x=1010$.
2) Let $x<0$, then $(x+x)^{2}+x-x=2020$, we get $4 x^{2}=2020, x= \pm \sqrt{505}$, but since $x<0$, then $x=-\sqrt{505}$.