\(\left|2x+1\right|=\dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}2x+1=\dfrac{3}{5}\\2x+1=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{5}-1\\2x=-\dfrac{3}{5}-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=-\dfrac{2}{5}\\2x=-\dfrac{8}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}:2\\x=-\dfrac{8}{5}:2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{5};-\dfrac{4}{5}\right\}\)
