Lời giải:
$(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020})x=\frac{2019}{1}+\frac{2018}{2}+....+\frac{2}{2018}+\frac{1}{2019}$
$=(\frac{2018}{2}+1)+(\frac{2017}{3}+1)+....+(\frac{2}{2018}+1)+(\frac{1}{2019}+1)+1$
$=\frac{2020}{2}+\frac{2020}{3}+....+\frac{2020}{2018}+\frac{2020}{2019}+\frac{2020}{2020}=2020(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2019}+\frac{1}{2020})$
$\Rightarrow x=2020$