Ta có: \(\left|x-\dfrac{1}{3}\right|=\left|2-3x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2-3x\\x-\dfrac{1}{3}=3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}-2+3x=0\\x-\dfrac{1}{3}-3x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{7}{3}=0\\-2x+\dfrac{5}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{7}{3}\\-2x=-\dfrac{5}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}:4=\dfrac{7}{12}\\x=-\dfrac{5}{3}:\left(-2\right)=\dfrac{-5}{3}\cdot\dfrac{1}{-2}=\dfrac{5}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{7}{12};\dfrac{5}{6}\right\}\)
