\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2021}{2022}\)
=>\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2021}{2022}\)
=>\(1-\dfrac{1}{x+2}=\dfrac{2021}{2022}\)
=>\(\dfrac{1}{x+2}=1-\dfrac{2021}{2022}=\dfrac{1}{2022}\)
=>x+2=2022
=>x=2020