a: \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)
=>\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}=\dfrac{-4}{10}=-\dfrac{2}{5}\)
=>\(x=-\dfrac{2}{5}:\dfrac{2}{3}=-\dfrac{3}{5}\)
b: \(\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{7}{10}\)
=>\(\dfrac{2}{3}x=\dfrac{7}{10}-\dfrac{1}{5}=\dfrac{1}{2}\)
=>\(x=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
c: \(\left(3\dfrac{4}{5}-2x\right)\cdot1\dfrac{1}{3}=5\dfrac{5}{7}\)
=>\(\left(\dfrac{19}{5}-2x\right)\cdot\dfrac{4}{3}=\dfrac{40}{7}\)
=>\(\dfrac{19}{5}-2x=\dfrac{40}{7}:\dfrac{4}{3}=\dfrac{40}{7}\cdot\dfrac{3}{4}=\dfrac{30}{7}\)
=>\(2x=\dfrac{19}{5}-\dfrac{30}{7}=\dfrac{-17}{35}\)
=>\(x=-\dfrac{17}{35}:2=-\dfrac{17}{70}\)
d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
