\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2020}{2021}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2020}{2021}\)
\\(1-\frac{1}{x+1}=\frac{2020}{2021}\)
\(\frac{1}{x+1}=1-\frac{2020}{2021}\)
\(\frac{1}{x+1}=\frac{1}{2021}\)
\(\Rightarrow x+1=2021\)
\(x=2021-1\)
\(x=2020\)
đk: \(x\ne\left\{0;-1\right\}\)
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2020}{2021}\)
\(\Leftrightarrow\frac{x}{x+1}=\frac{2020}{2021}\)
\(\Leftrightarrow2021x=2020x+2020\)
\(\Rightarrow x=2020\)