|x + $\dfrac{3}{3}$ | - $\dfrac{1}{3}$ = 0
|x + $\dfrac{3}{3}$ | = 0 + $\dfrac{1}{3}$
|x + $\dfrac{3}{3}$ | = $\dfrac{1}{3}$
⇒\(\left[ \begin{array}{l}x+\dfrac{3}{3}\ =\dfrac{1}{3}\\x+\dfrac{3}{3}\ = \dfrac{-1}{3}\ \end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=\dfrac{1}{3}\ - \dfrac{3}{3}\ \\x= \dfrac{-1}{3}-\dfrac{3}{3}\ \end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=\dfrac{-2}{3}\ \\x= \dfrac{-4}{3}\ \end{array} \right.\)
Vậy x= \(\left[ \begin{array}{l}x=\dfrac{-2}{3}\ \\x= \dfrac{-4}{3}\ \end{array} \right.\)
