` | 1/2 . x - 1/6 | = 1/3`
`=>` \(\left[ \begin{array}{l}1/2 . x - 1/6 = 1/3\\1/2 . x - 1/6 = - 1/3\end{array} \right.\)
`TH1 : 1/2 . x - 1/6 = 1/3`
`=> 1/2 . x = 1/2`
`=> x = 1`
`TH2 : 1/2 . x - 1/6 = - 1/3`
`=> 1/2 . x = -1/6`
`=> x = -1/3`
Vậy \(\left[ \begin{array}{l}x=1\\x=-1/3\end{array} \right.\)