Ta có: \(\left|2x-\frac{1}{7}\right|+\frac{2}{7}=\frac{7}{5}\)
\(\Leftrightarrow\left|2x-\frac{1}{7}\right|=\frac{7}{5}-\frac{2}{7}=\frac{49}{35}-\frac{10}{35}=\frac{39}{35}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{1}{7}=\frac{39}{35}\\2x-\frac{1}{7}=\frac{-39}{35}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{39}{35}+\frac{1}{7}=\frac{39}{35}+\frac{5}{35}=\frac{44}{35}\\2x=\frac{-39}{35}+\frac{1}{7}=\frac{-39}{35}+\frac{5}{35}=\frac{-34}{35}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{44}{35}:2=\frac{44}{35}\cdot\frac{1}{2}=\frac{22}{35}\\x=\frac{-34}{35}:2=\frac{-34}{35}\cdot\frac{1}{2}=-\frac{17}{35}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-17}{35};\frac{22}{35}\right\}\)
