\(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2.\left(1-\frac{1}{x+1}\right)\)
\(=2-\frac{2}{x+1}\) mà \(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)
=> \(2-\frac{2}{x+1}=1\frac{2008}{2010}=>\frac{2}{x+1}=\frac{2}{2010}=>x+1=2010=>x=2009\)
đúng cái nhé
