$\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10} + .... + \dfrac{2}{x.(x+1)} = \dfrac{2007}{2009}$
$⇔ \dfrac{2}{6} + \dfrac{2}{12} + \dfrac{2}{20} + .... + \dfrac{2}{x.(x+1)} = \dfrac{2007}{2009}$
$⇔ 2 . (\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + .... + \dfrac{1}{x.(x+1)}) = \dfrac{2007}{2009}$
$⇔ 2 . (\dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + .... + \dfrac{1}{x.(x+1)}) = \dfrac{2007}{2009}$
$⇔ 2 . ( \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ...... + \dfrac{1}{x} - \dfrac{1}{x+1}) = \dfrac{2007}{2009}$
$⇔ 2 . (\dfrac{1}{2} - \dfrac{1}{x+1}) = \dfrac{2007}{2009}$
$⇔ 1 - \dfrac{2}{x+1} = \dfrac{2007}{2009}$
$⇔ \dfrac{2}{x+1} = \dfrac{2}{2009}$
$⇔ x+1 = 2009$
$⇔ x = 2008$
Vậy $x=2008$
