\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2007}{2009}\)
\(\dfrac{2}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}\right)=\dfrac{2007}{2009}\)
\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2007}{2009}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2007}{2009}:2\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2007}{4018}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2007}{4018}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2007}{4018}\)
\(\dfrac{1}{x+1}=\dfrac{2009}{4018}-\dfrac{2007}{4018}\)
\(\dfrac{1}{x+1}=\dfrac{2}{4018}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(x=2009-1\)
\(x=2008\).
Vậy x = 2008 .
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