Đáp án đúng: D
Giải chi tiết:\(1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \ldots + \frac{1}{{x\left( {x + 1} \right):2}} = 1\frac{{2019}}{{2021}}\)
\(\begin{array}{l}1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \ldots + \frac{1}{{x\left( {x + 1} \right):2}} = 1\frac{{2019}}{{2021}}\\\frac{1}{2} \cdot \left( {1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{{10}} + \ldots + \frac{1}{{x\left( {x + 1} \right):2}}} \right) = \frac{{4040}}{{2021}} \cdot \frac{1}{2}\\\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \ldots + \frac{1}{{x\left( {x + 1} \right)}} = \frac{{2020}}{{2021}}\\\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + \ldots + \frac{1}{{x\left( {x + 1} \right)}} = \frac{{2020}}{{2021}}\\1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{x} - \frac{1}{{x + 1}} = \frac{{2020}}{{2021}}\\1 - \frac{1}{{x + 1}} = \frac{{2020}}{{2021}}\\\,\,\,\,\,\,\frac{1}{{x + 1}} = 1 - \frac{{2020}}{{2021}}\\\,\,\,\,\,\,\frac{1}{{x + 1}} = \frac{1}{{2021}}\\\,\,\,\,\,\,x + 1 = 2021\\\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\, = 2021 - 1\\\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\, = 2020.\end{array}\)
Vậy \(x = 2020.\)
Chọn D.
