Question

[Nguyễn Ngọc Huyền]

Tìm X : \(1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2009}{2011}\)

Answer 1

[Lam Vu Thien Phuc]

=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)

=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)

=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)

=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)

=> x + 1 = 2011

=> x = 2010