giai phương trình (x+ $\frac{1}{6}$ )+(x+$\frac{1}{12}$ )+…+(x +$\frac{1}{72}$ ) = $\frac{133}{18}$

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

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