Lời giải:
$\frac{x+2}{2020}+\frac{x+2}{2020}=\frac{x+2019}{3}+\frac{x+2020}{2}$
$\frac{x+2}{2020}+1+\frac{x+2}{2020}+2=\frac{x+2019}{3}+1+\frac{x+2020}{2}+1$
$\frac{x+2022}{2020}+\frac{x+2022}{2020}=\frac{x+2022}{3}+\frac{x+2022}{2}$
$(x+2022)(\frac{1}{2020}+\frac{1}{2020}-\frac{1}{3}-\frac{1}{2})=0$
Dễ thấy $\frac{1}{2020}+\frac{1}{2020}-\frac{1}{3}-\frac{1}{2}<0$
Do đó: $x+2022=0$
$\Rightarrow x=-2022$