\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\)
\(\Leftrightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{14}{5}\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{14}{5}-\dfrac{4}{5}\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-5}{3}\end{matrix}\right.\)
Vậy..............
\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \left | \left ( - 3,2 \right ) + \frac{2}{5}\right |\)
\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \left | - \frac{14}{5} \right |\)
\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \frac{14}{5} \)
\(\left | x - \frac{1}{3} \right | = 2\)
* \(x - \frac{1}{3}= 2\)
x = 2 + \( \frac{1}{3}\)
\(x = \frac{7}{3}\)
* \(x - \frac{1}{3}= - 2\)
\(x = - 2 + \frac{1}{3}\)
\(x = - \frac{5}{3}\)
Vậy x = \(x = \frac{7}{3}; x = - \frac{5}{3}\)
