3\(-|\dfrac{1}{3}-x|=\dfrac{5}{17}\)

\(|\dfrac{1}{3}-x|\) \(=\dfrac{5}{17}-3\)

\(|\dfrac{1}{3}-x|\)\(=\dfrac{5}{17}-\dfrac{51}{17}\)

\(|\dfrac{1}{3}-x|\)\(=\dfrac{46}{17}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{46}{17}\\\dfrac{1}{3}-x=\dfrac{-46}{17}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-\dfrac{46}{17}\\x=\dfrac{1}{3}-\dfrac{-46}{17}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{-46}{17}\\x=\dfrac{1}{3}=\dfrac{46}{17}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{51}+\dfrac{-138}{51}\\x=\dfrac{17}{51}+\dfrac{138}{51}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-121}{51}\\x=\dfrac{155}{51}\end{matrix}\right.\)

Vậy \(x=\left\{\dfrac{-121}{51};\dfrac{155}{51}\right\}\)