\(3\left(\dfrac{1}{5}-x\right)=-\dfrac{9}{5}\)

⇔\(\left(\dfrac{1}{5}-x\right)=-\dfrac{9}{5}:3\)

⇔​​\(\left(\dfrac{1}{5}-x\right)=-\dfrac{9}{5}.\dfrac{1}{3}\)​​

⇔\(\left(\dfrac{1}{5}-x\right)=-\dfrac{3}{5}\)

⇔\(x=\dfrac{1}{5}-\left(-\dfrac{3}{5}\right)\)

⇔\(x=\dfrac{1}{5}+\dfrac{3}{5}\)

⇔\(x=\dfrac{4}{5}\)

Vậy \(x=\dfrac{4}{5}\)