\(\dfrac{-2}{5}+\dfrac{4}{5}x=\dfrac{3}{5}\\ \dfrac{4}{5}x=\dfrac{3}{5}+\dfrac{2}{5}\\ \dfrac{4}{5}x=1\\ x=1:\dfrac{4}{5}\\ x=\dfrac{5}{4}\)
\(\dfrac{4}{5}x=\dfrac{3}{5}-\dfrac{-2}{5}\)
\(\dfrac{4}{5}x=1\)
-> x= \(\dfrac{5}{4}\)
Ta có: \(\dfrac{-2}{5}+\dfrac{4}{5}x=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{4}{5}x=\dfrac{3}{5}+\dfrac{2}{5}=1\)
hay \(x=\dfrac{5}{4}\)
Vậy: \(x=\dfrac{5}{4}\)