\(a)\dfrac{3}{2}-x=\dfrac{1}{2}-7\\ \dfrac{3}{2}-x=-\dfrac{13}{2}\\ x=\dfrac{3}{2}+\dfrac{13}{2}\\ x=8\\ b)\dfrac{4}{5}x-\dfrac{1}{2}=-\dfrac{3}{4}\\ \dfrac{4}{5}x=-\dfrac{3}{4}+\dfrac{1}{2}\\ \dfrac{4}{5}x=-\dfrac{1}{4}\\ x=-\dfrac{1}{4}:\dfrac{4}{5}\\ x=-\dfrac{5}{16}\\ c)\dfrac{5}{6}+\dfrac{1}{6}:x=-\dfrac{1}{3}\\ \dfrac{1}{6}:x=-\dfrac{1}{3}-\dfrac{5}{6}=-\dfrac{7}{6}\\ x=\dfrac{1}{6}:-\dfrac{7}{6}\\ x=-\dfrac{1}{7}\\ d)-\dfrac{2}{3}-\left(2x+\dfrac{1}{2}\right)=-\dfrac{3}{2}\\ 2x+\dfrac{1}{2}=-\dfrac{2}{3}+\dfrac{3}{2}\\ 2x+\dfrac{1}{2}=\dfrac{5}{6}\\ 2x=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}:2=\dfrac{1}{6}\\ e)\left(\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}+...+\dfrac{1}{29\cdot30}\right)\cdot120+x:\dfrac{1}{3}=-4\\ \left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right)\cdot120+x:\dfrac{1}{3}=-4\\ \left(\dfrac{1}{24}-\dfrac{1}{30}\right)\cdot120+x:\dfrac{1}{3}=-4\\ x:\dfrac{1}{3}=-4-1\\ x:\dfrac{1}{3}=-5\\ x=-5\cdot\dfrac{1}{3}=-\dfrac{5}{3}\)
