Ta có : $|x+\dfrac{3}{4}|-\dfrac{1}{3} = 0$
$⇔|x+\dfrac{3}{4}|=\dfrac{1}{3}$
$⇔ \left[ \begin{array}{l}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=\dfrac{-1}{3}\end{array} \right. $
$⇔ \left[ \begin{array}{l}x=\dfrac{-5}{12}\\x=\dfrac{-13}{12}\end{array} \right. $