\(a,\\ \left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\\ \Leftrightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}=3\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-x=3\\\dfrac{2}{5}-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}-3=-\dfrac{13}{5}\\x=\dfrac{2}{5}-\left(-3\right)=\dfrac{17}{5}\end{matrix}\right.\)
\(b,\left|x-\dfrac{1}{3}\right|=2\dfrac{1}{5}=\dfrac{11}{5}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{11}{5}\\x-\dfrac{1}{3}=-\dfrac{11}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{5}+\dfrac{1}{3}=\dfrac{38}{15}\\x=-\dfrac{11}{5}+\dfrac{1}{3}=-\dfrac{28}{15}\end{matrix}\right.\)
