$\frac{1}{2}$+$\frac{1}{6}$+$\frac{1}{12}$+....+$\frac{1}{x.(x+1)}$ = $\frac{2020}{2021}$
$\frac{1}{1.2}$+$\frac{1}{2.3}$+$\frac{1}{3.4}$+....+$\frac{1}{x.(x+1)}$ = $\frac{2020}{2021}$
$1-\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+....+$\frac{1}{x}$-$\frac{1}{x+1}$= $\frac{2020}{2021}$
$1-\frac{1}{x+1}$=$\frac{2020}{2021}$
$\frac{1}{x+1}$=$1-\frac{2020}{2021}$
$\frac{1}{x+1}$=$\frac{1}{2021}$
⇒x+1=2021
⇒x=2020
Vậy x= 2020
