Tìm số tự nhiên x biết 1/3 +1/6 +1/10 +...+ 1/x(x+1) : 2 = 2019/2021

* Hướng dẫn giải

Đáp án cần chọn là: C

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