$\dfrac{1}{x} + \dfrac{1}{x+1} = \dfrac{5}{6}$
$⇔ \dfrac{x+1}{x.(x+1)} + \dfrac{x}{x.(x+1)} = \dfrac{5}{6}$
$⇔ \dfrac{2x+1}{x.(x+1)} = \dfrac{5}{6}$
$⇔ 6(2x+1) = 5x(x+1)$
$⇔ 12x + 6 - 5x^2 - 5x=0$
$⇔ 5x^2 - 7x - 6=0$
$⇔5x^2 + 3x -10x - 6=0$
$⇔ x(5x + 3) - 2(5x+3) = 0$
$⇔ (x-2)(5x+3)=0$
$⇒$ \(\left[ \begin{array}{l}x=2(TM)\\x=\dfrac{-3}{5}(TM)\end{array} \right.\)
Vậy $x$ $∈$ `{2;\frac{-3}{5}}`
