Ta có: \(|x-3|-x=2x-\frac{5}{3}\)
\(\Leftrightarrow|x-3|=3x-\frac{5}{3}\left(1\right)\)
Mà \(|x-3|\ge0;\forall x\)
\(\Rightarrow3x-\frac{5}{3}\ge0\)
\(\Rightarrow x\ge\frac{5}{9}\)
Từ (1) \(\Rightarrow\orbr{\begin{cases}x-3=3x-\frac{5}{3}\\x-3=\frac{5}{3}-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2x=\frac{4}{3}\\4x=\frac{14}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}< \frac{5}{9}\left(loai\right)\\x=\frac{7}{6}>\frac{5}{9}\left(chon\right)\end{cases}}\)
Vậy \(x=\frac{7}{6}\)
