\(2\left|-x+\dfrac{5}{6}\right|-\dfrac{1}{3}=2\\ \Rightarrow2\left|-x+\dfrac{5}{6}\right|=\dfrac{7}{3}\\ \Rightarrow\left|-x+\dfrac{5}{6}\right|=\dfrac{7}{6}\\ \Rightarrow\left[{}\begin{matrix}-x+\dfrac{5}{6}=-\dfrac{7}{6}\\-x+\dfrac{5}{6}=\dfrac{7}{6}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\left|-x+\dfrac{5}{6}\right|-\dfrac{1}{3}=2\)
<=> \(\left[{}\begin{matrix}-x+\dfrac{5}{6}-\dfrac{1}{3}=2\left(x\le\dfrac{5}{6}\right)\\x-\dfrac{5}{6}-\dfrac{1}{3}=2\left(x>\dfrac{5}{6}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\left(2-\dfrac{5}{6}+\dfrac{1}{2}\right)\\x=2+\dfrac{5}{6}+\dfrac{1}{3}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\left(TM\right)\\x=\dfrac{19}{6}\left(TM\right)\end{matrix}\right.\)
