Bạn tham khảo :
Ta có :
$|x+\dfrac{4}{5}| = \dfrac{1}{5}$
⇒ \(\left[ \begin{array}{l}x+\dfrac{4}{5}=\dfrac{1}{5}\\x+\dfrac{4}{5}=-\dfrac{1}{5}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x= \dfrac{-3}{5}\\x= -1\end{array} \right.\)
Vậy $x∈${$\dfrac{-3}{5} ; (-1)$}