$(x+\dfrac{1}{5})^2 + \dfrac{18}{25} = \dfrac{27}{25}$
$⇔ (x+\dfrac{1}{5})^2 = \dfrac{9}{25}$
$⇒$ \(\left[ \begin{array}{l}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5} = \dfrac{-3}{5}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{array} \right.\)
Vậy $x$ ∈ {$\dfrac{-4}{5};\dfrac{2}{5}$}