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

* Hướng dẫn giải

$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\mathrm{...}+\frac{1}{x\left(x+1\right):2}=\frac{2019}{2021}$

$2.\left[\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{x\left(x+1\right)}\right]=\frac{2019}{2021}$

$2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2019}{2021}$

$2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2021}$

$1-\frac{2}{x+1}=\frac{2019}{2021}$

$\frac{2}{x+1}=1-\frac{2019}{2021}$

$\frac{2}{x+1}=\frac{2}{2021}$