\(\left(\dfrac{2}{3}+x\right)\left(\dfrac{3}{5}-2x\right)=10\)

\(\Leftrightarrow\dfrac{2}{3}+x;\dfrac{3}{5}-2x\inƯ\left(10\right)\)

\(Ư\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

\(\Leftrightarrow\dfrac{2}{3}+x=1\Rightarrow x=\dfrac{1}{3}\)

\(\dfrac{3}{5}-2x=10\Rightarrow2x=\dfrac{-47}{5}\Rightarrow x=\dfrac{-47}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-1\Rightarrow x=\dfrac{-5}{3}\)

\(\dfrac{3}{5}-2x=-10\Rightarrow2x=\dfrac{53}{5}\Rightarrow x=\dfrac{53}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=2\Rightarrow x=\dfrac{4}{3}\)

\(\dfrac{3}{5}-2x=5\Rightarrow2x=\dfrac{-22}{5}\Rightarrow x=\dfrac{-11}{5}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-2\Rightarrow x=\dfrac{-8}{3}\)

\(\dfrac{3}{5}-2x=-5\Rightarrow2x=\dfrac{28}{5}\Rightarrow x=\dfrac{14}{5}\)

\(\Leftrightarrow\dfrac{2}{3}+x=5\Rightarrow x=\dfrac{13}{3}\)

\(\dfrac{3}{5}-2x=2\Rightarrow2x=\dfrac{-7}{5}\Rightarrow x=\dfrac{-7}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-5\Rightarrow x=\dfrac{-17}{3}\)

\(\dfrac{3}{5}-2x=-2\Rightarrow2x=\dfrac{13}{5}\Rightarrow x=\dfrac{13}{10}\)

\(\Leftrightarrow\dfrac{2}{3}+x=10\Rightarrow x=\dfrac{28}{3}\)

\(\dfrac{3}{5}-2x=1\Rightarrow2x=\dfrac{-2}{5}\Rightarrow x=\dfrac{-1}{5}\)

\(\Leftrightarrow\dfrac{2}{3}+x=-10\Rightarrow x=\dfrac{-32}{3}\)

\(\dfrac{3}{5}-2x=-1\Rightarrow2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\)