=> \(\dfrac{1}{2}.x-\dfrac{1}{3}.x=\dfrac{1}{6}-\dfrac{2}{3}\)
\(x.\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{1}{6}+\dfrac{-4}{6}\)
\(x.\left(\dfrac{3}{6}+\dfrac{-2}{6}\right)=\dfrac{-1}{2}\)
\(x.\dfrac{1}{6}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{1}{6}\)
\(x=\dfrac{-1}{2}.6\)
\(x=-3\)
Vậy x= -3
\(\dfrac{1}{2}x+\dfrac{2}{3}=\dfrac{1}{3}x+\dfrac{1}{6}\)
\(\dfrac{1}{2}x-\dfrac{1}{3}x=\dfrac{1}{6}-\dfrac{2}{3}\)
\(x\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{1}{6}-\dfrac{4}{6}\)
\(x\left(\dfrac{3}{6}-\dfrac{2}{6}\right)=-\dfrac{1}{2}\)
\(x\cdot\dfrac{1}{6}=-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}:\dfrac{1}{6}\)
\(x=-\dfrac{1}{2}\cdot6\)
\(x=-3\)
Vậy \(x=-3\).