Để: $\dfrac{5-2x}{6}<\dfrac{3+x}{2}$
`<=>`$\dfrac{5-2x}{6}-\dfrac{3+x}{2}<0$
`<=>`$\dfrac{(5-2x)-3(3+x)}{6}<0$
`<=>`$\dfrac{5-2x-9-3x}{6}<0$
`<=>`$\dfrac{-4-5x}{6}<0$
`<=>`$-4-5x<0 ($vì: $6>0)$
`<=>`$-5x<4$
`<=>`$x>\dfrac{-4}{5}$
Vậy $x>\dfrac{-4}{5}$ thì $\dfrac{5-2x}{6}<\dfrac{3+x}{2}$