Question

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

*Tìm x*

Answer 1

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{2008}{2009}\)

\(\Rightarrow1-\dfrac{1}{x}=\dfrac{2008}{2009}\)

\(\Rightarrow x=2009\)

Answer 2

<=> \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

<=> \(1-\dfrac{1}{x+1}=\dfrac{2008}{2009}=\dfrac{x}{x+1}=\dfrac{2008}{2009}\)

=> x = 2008