Trường hợp 1:
x + \(\dfrac{1}{3}\) = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{3}{6}\)
x = \(\dfrac{1}{2}\)
Trường hợp 2:
x + \(\dfrac{1}{3}\)= \(\dfrac{-5}{6}\)
x = \(\dfrac{-5}{6}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{-7}{6}\)
Vậy x ∈ { \(\dfrac{1}{2}\);\(\dfrac{-7}{6}\)}