Ta có: $\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}=(\frac{1}{199}+1)+(\frac{2}{198}+...+(\frac{199}{1}+1)-199$$=200(\frac{1}{199}+\frac{1}{198}+...\frac{1}{2}+1)-199$
$=200(\frac{1}{200}+\frac{1}{199}+...+\frac{1}{2}+1-\frac{1}{200})-199$
$=200(\frac{1}{200}+\frac{1}{199}+...+\frac{1}{2})+200-1-199$
$=200(\frac{1}{200}+\frac{1}{199}+...+\frac{1}{2})$
Khi đó pt đã cho tương đương:
$(x-20)*\frac{1}{200}=\frac{1}{2000}=>x=\frac{201}{10}$