Question

[Hyu Hinata]

Tìm x : \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{1999}{2001}\)

Answer 1

[Trần Thị Loan]

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{1999}{2001}\)

=> \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1999}{2001}\)

=> \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1999}{2001}\)

=> \(1-\frac{1}{x+1}=\frac{4000}{2001}:2\) =>\(1-\frac{1}{x+1}=\frac{2000}{2001}\) =>   \(\frac{1}{x+1}=\frac{1}{2001}\) => x+ 1 = 2001 => x = 2000

Vậy...........

Answer 2

[Nguyễn Ngọc Quý]

Mình thấy \(1+\frac{1}{3}+\frac{1}{6}+.......\) mà sao cô ghi 2 nhỉ