$( x + \dfrac{1}{6} ) + ( x + \dfrac{1}{12} ) + ... + ( x + \dfrac{1}{72} ) = \dfrac{133}{18}$
$⇔ 12x + ( \dfrac{1}{6} + \dfrac{1}{12} + ... + \dfrac{1}{72} ) = \dfrac{133}{18}$
$⇔ 12x + ( \dfrac{1}{2.3} + \dfrac{1}{3.4} + ... + \dfrac{1}{8.9} ) = \dfrac{133}{18}$
$⇔ 12x + ( \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{8} - \dfrac{1}{9} ) = \dfrac{133}{18}$
$⇔ 12x + ( \dfrac{1}{2} - \dfrac{1}{9} ) = \dfrac{133}{18}$
$⇔ 12x + \dfrac{7}{18} = \dfrac{133}{18}$
$⇔ x = \dfrac{7}{12}$
