Bạn giải cụ thể đi ạ
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1009}{1010}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1009}{1010}\)
=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1009}{1010}:2\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1009}{2020}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2020}\)
=> \(\frac{1}{x+1}=\frac{1}{2020}\)
=> \(x+1=2020\)
=> x = 2019
Vậy x = 2019