\(3\left(x-\dfrac{1}{6}\right)-2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{2}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
3 x \(\left(x-\dfrac{1}{6}\right)\) - 2 x \(\left(x-\dfrac{1}{3}\right)\) = \(\dfrac{2}{3}\)
3 x \(x\) - \(3\) x \(\dfrac{1}{6}\) - 2 x \(x\) + 2 x \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(x\) - \(\dfrac{1}{2}\) + \(\dfrac{2}{3}\) = \(\dfrac{2}{3}\)
\(x-\dfrac{1}{2}\) = \(\dfrac{2}{3}-\dfrac{2}{3}\)
\(x-\dfrac{1}{2}\) = 0
\(x=\dfrac{1}{2}\)