\(\left|\dfrac{1}{5}x+\dfrac{2}{5}\right|=\left|\dfrac{2}{3}x-\dfrac{1}{3}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{5}x+\dfrac{2}{5}=\dfrac{2}{3}x-\dfrac{1}{3}\\\dfrac{1}{5}x+\dfrac{2}{5}=\dfrac{-2}{3}x+\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{-7}{15}=\dfrac{-11}{15}\\x\cdot\dfrac{13}{15}=\dfrac{-1}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{7}\\x=\dfrac{-1}{13}\end{matrix}\right.\)