\(\Rightarrow\)\(\left[3x+\dfrac{3}{2}+3-5x\right].\dfrac{1}{2}=-x:\dfrac{31}{5}\)
\(\Rightarrow\)\(\left[-2x+\dfrac{9}{2}\right].\dfrac{1}{2}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(\dfrac{-2x}{2}+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(-x+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5}{31}+x\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5+31x}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x\left(-5+31\right)}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x.26}{31}\)
\(\Rightarrow\dfrac{9}{4}.31=x.26\)\(\Rightarrow\dfrac{279}{4}=x.26\)
\(\Rightarrow x=\dfrac{279}{4}:26=\dfrac{279}{4}.\dfrac{1}{26}\)
\(\Rightarrow x=\dfrac{279}{104}\)
Vậy x=\(\dfrac{279}{104}\)
