\(\left(x-\dfrac{3}{5}\right)\left(x+\dfrac{3}{8}\right)>0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>0\\x+\dfrac{3}{8}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-\dfrac{3}{5}< 0\\x+\dfrac{3}{8}< 0\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x-\dfrac{3}{5}>0\\x+\dfrac{3}{8}>0\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}x>\dfrac{3}{5}\\x>-\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow x>\dfrac{3}{5}\)
Với \(\left\{{}\begin{matrix}x-\dfrac{3}{5}< 0\\x+\dfrac{3}{8}< 0\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}x< \dfrac{3}{5}\\x< -\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow x< -\dfrac{3}{8}\)
Vậy x \(\in Q\) thỏa mãn \(x< -\dfrac{3}{8}\) hoặc \(x>\dfrac{3}{5}\) \(\left(đpcm\right)\)