\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\\ \left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\\ \left|x-\frac{1}{3}\right|=2 \\ \Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-\frac{5}{3}\end{matrix}\right.\)
Vậy...
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
=> \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-\frac{16}{5}\right)+\frac{2}{5}\right|\)
=> \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-\frac{14}{5}\right|=\frac{14}{5}\)
=> \(\left[{}\begin{matrix}x-\frac{1}{3}=\frac{14}{5}\\x-\frac{1}{3}=-\frac{14}{5}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{14}{5}+\frac{1}{3}\\x=-\frac{14}{5}+\frac{1}{3}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{47}{15}\\x=-\frac{37}{15}\end{matrix}\right.\)
Vậy:..................................
P/s: Ko chắc!