\(\Leftrightarrow\dfrac{1}{5}-\dfrac{3}{2}x=\pm\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\\\dfrac{1}{5}-\dfrac{3}{2}x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{2}x=\dfrac{3}{2}-\dfrac{1}{5}=\dfrac{13}{10}\\-\dfrac{3}{2}x=-\dfrac{3}{2}-\dfrac{1}{5}=-\dfrac{17}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{10}:\left(-\dfrac{3}{2}\right)=-\dfrac{13}{15}\\x=-\dfrac{17}{10}:\left(-\dfrac{3}{2}\right)=\dfrac{17}{15}\end{matrix}\right.\)
Vậy \(x=-\dfrac{13}{15};x=\dfrac{17}{15}\)