Giải:
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(C=\left|x-\dfrac{2}{5}\right|+\left|x-\dfrac{3}{5}\right|=\left|x-\dfrac{2}{5}\right|+\left|\dfrac{3}{5}-x\right|\)
\(\ge\left|x-\dfrac{2}{5}+\dfrac{3}{5}-x\right|=\left|\dfrac{1}{5}\right|=\dfrac{1}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{2}{5}\ge0\\\frac{3}{5}-x\ge0\end{cases}}\) \(\Leftrightarrow\dfrac{2}{5}\le x\le\dfrac{3}{5}\)
Vậy \(C_{min}=\dfrac{1}{5}\) \(\Leftrightarrow\dfrac{2}{5}\le x\le\dfrac{3}{5}\)