\(\left(\dfrac{2}{5}-3x\right)^2-\dfrac{1}{5}=\dfrac{4}{25}\)
=>\(\left(3x-\dfrac{2}{5}\right)^2=\dfrac{4}{25}+\dfrac{1}{5}=\dfrac{9}{25}\)
=>\(\left[{}\begin{matrix}3x-\dfrac{2}{5}=\dfrac{3}{5}\\3x-\dfrac{2}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{3}{5}+\dfrac{2}{5}=1\\3x=-\dfrac{3}{5}+\dfrac{2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{15}\end{matrix}\right.\)
\(\left(\frac{2}{5} - 3x\right)^2 - \frac{1}{5} = \frac{4}{25} \Rightarrow \left(3x - \frac{2}{5}\right)^2 = \frac{4}{25} + \frac{1}{5} = \frac{9}{25} \Rightarrow \begin{cases} 3x - \frac{2}{5} = \frac{3}{5} \\ 3x - \frac{2}{5} = -\frac{3}{5} \end{cases} \Leftrightarrow \begin{cases} 3x = \frac{3}{5} + \frac{2}{5} = 1 \\ 3x = -\frac{3}{5} + \frac{2}{5} = -\frac{1}{5} \end{cases} \Rightarrow \begin{cases} x = \frac{1}{3} \\ x = -\frac{1}{15} \end{cases}\)
