Ta có: \(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(\Leftrightarrow2\left(2x-3\right)-9=5-3x-2\)
\(\Leftrightarrow4x-15=-3x+3\)
\(\Leftrightarrow4x+3x=3+15\)
\(\Leftrightarrow x=\dfrac{18}{7}\)
\(\Leftrightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}\)
\(\Leftrightarrow\dfrac{4x-6}{6}-\dfrac{5-3x}{6}=\dfrac{7}{6}\)
\(\Leftrightarrow4x-6-5+3x=7\)
\(\Leftrightarrow7x=7+5+6=18\)
\(\Leftrightarrow x=\dfrac{18}{7}\)
Vậy ...