\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{1\cdot2}{2}}+\dfrac{1}{\dfrac{2\cdot3}{2}}+\dfrac{1}{\dfrac{3\cdot4}{2}}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{4040}{2021}\)
\(\Leftrightarrow\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4040}{2021}\)
\(\Leftrightarrow2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4040}{2021}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow2021x=2020x+2020\Leftrightarrow x=2020\)
Vậy S = {2020}
