Tìm số tự nhiên $x$ biết $\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{{10}} + ... + \dfrac{1}{{x\left( {x + 1} \right):2}} = \dfrac{{2019}}{{2021}}$

$\begin{array}{l}\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{{10}} + ... + \dfrac{1}{{x\left( {x + 1} \right):2}} = \dfrac{{2019}}{{2021}}\\2.\left[ {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{x(x + 1)}}} \right] = \dfrac{{2019}}{{2021}}\\2.\left( {\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right) = \dfrac{{2019}}{{2021}}\\2.\left( {\dfrac{1}{2} - \dfrac{1}{{x + 1}}} \right) = \dfrac{{2019}}{{2021}}\\1 - \dfrac{2}{{x + 1}} = \dfrac{{2019}}{{2021}}\\\dfrac{2}{{x + 1}} = 1 - \dfrac{{2019}}{{2021}}\\\dfrac{2}{{x + 1}} = \dfrac{2}{{2021}}\\x + 1 = 2021\\x = 2020\end{array}$