a) \(\dfrac{2}{3}\cdot x+\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}\cdot x=\dfrac{1}{10}-\dfrac{1}{2}\)
\(\dfrac{2}{3}\cdot x=\dfrac{1}{10}-\dfrac{5}{10}\)
\(\dfrac{2}{3}\cdot x=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}:\dfrac{2}{3}\)
\(x=\dfrac{-2}{5}\cdot\dfrac{3}{2}\)
\(x=\dfrac{-3}{5}\)
Vậy \(x=\dfrac{-3}{5}\).
b) \(\dfrac{2}{3}\cdot x+\dfrac{1}{5}=\dfrac{7}{10}\)
\(\dfrac{2}{3}\cdot x=\dfrac{7}{10}-\dfrac{1}{5}\)
\(\dfrac{2}{3}\cdot x=\dfrac{7}{10}-\dfrac{2}{10}\)
\(\dfrac{2}{3}\cdot x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}:\dfrac{2}{3}\)
\(x=\dfrac{1}{2}\cdot\dfrac{3}{2}\)
\(x=\dfrac{3}{4}\)
Vậy \(x=\dfrac{3}{4}\).
c) \(\left(3\dfrac{4}{5}-2\cdot x\right)\cdot1\dfrac{1}{3}=5\dfrac{5}{7}\)
\(\left(\dfrac{19}{5}-2\cdot x\right)\cdot\dfrac{4}{3}=\dfrac{40}{7}\)
\(\dfrac{19}{5}-2\cdot x=\dfrac{40}{7}:\dfrac{4}{3}\)
\(\dfrac{19}{5}-2\cdot x=\dfrac{40}{7}\cdot\dfrac{3}{4}\)
\(\dfrac{19}{5}-2\cdot x=\dfrac{30}{7}\)
\(2\cdot x=\dfrac{19}{5}-\dfrac{30}{7}\)
\(2\cdot x=\dfrac{133}{35}-\dfrac{150}{35}\)
\(2\cdot x=\dfrac{-17}{35}\)
\(x=\dfrac{-17}{35}:2\)
\(x=\dfrac{-17}{35}\cdot\dfrac{1}{2}\)
\(x=\dfrac{-17}{70}\)
Vậy \(x=\dfrac{-17}{70}\).
d) \(\dfrac{x}{7}=\dfrac{6}{-21}\)
\(\dfrac{x}{7}=\dfrac{-6}{21}\)
\(x=\dfrac{-6\cdot7}{21}\)
\(x=-2\)
Vậy \(x=-2\).
