\(\dfrac{2}{3}x-\dfrac{3}{2}\left(x-\dfrac{1}{2}\right)=\dfrac{5}{12}\)
\(\dfrac{2}{3}x-\dfrac{3}{2}x-\dfrac{3}{2}.\dfrac{1}{2}=\dfrac{5}{12}\)
\(\dfrac{2}{3}x-\dfrac{3}{2}x-\dfrac{3}{4}=\dfrac{5}{12}\)
\(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{7}{6}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{7}{6}\)
\(x.\dfrac{-5}{6}=\dfrac{7}{6}\)
\(x=\dfrac{7}{6}:\dfrac{-5}{6}\)
\(x=-\dfrac{7}{5}\)
Ta có: \(\dfrac{2}{3}x-\dfrac{3}{2}\left(x-\dfrac{1}{2}\right)=\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{3}{2}x+\dfrac{3}{4}=\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{-5}{6}x=\dfrac{5}{12}-\dfrac{3}{4}=\dfrac{5}{12}-\dfrac{9}{12}=\dfrac{-1}{3}\)
hay \(x=\dfrac{-1}{3}:\dfrac{-5}{6}=\dfrac{1}{3}\cdot\dfrac{6}{5}=\dfrac{2}{5}\)