\(\left|x-\frac{1}{3}\right|=\left|2-3x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=2-3x\\x-\frac{1}{3}=3x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{7}{3}\\2x=-\frac{5}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{12}\\x=-\frac{5}{6}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{7}{12},-\frac{5}{6}\right\}\)
@Akai Haruma cô xem hộ em có thiếu TH không ạ ?
\(\left|x-\frac{1}{3}\right|=\left|2-3x\right|\)
=> \(\left[{}\begin{matrix}x-\frac{1}{3}=2-3x\\x-\frac{1}{3}=3x-2\end{matrix}\right.\) => \(\left[{}\begin{matrix}x+3x=2+\frac{1}{3}\\3x-x=\left(-2\right)+\frac{1}{3}\end{matrix}\right.\) => \(\left[{}\begin{matrix}4x=\frac{7}{3}\\2x=-\frac{5}{3}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{7}{3}:4\\x=\left(-\frac{5}{3}\right):2\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{7}{12}\\x=-\frac{5}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{7}{12};-\frac{5}{6}\right\}.\)
Chúc bạn học tốt!