a) \(\left(x-\dfrac{3}{5}\right)\left(x+\dfrac{3}{8}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{5}>0\\x+\dfrac{3}{8}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{5}< 0\\x+\dfrac{3}{8}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{5}\\x>-\dfrac{3}{8}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{5}\\x< -\dfrac{3}{8}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{3}{5}\\x< -\dfrac{3}{8}\end{matrix}\right.\)
Vậy ...
b) \(\left(2x+\dfrac{3}{2}\right):\left(2x-\dfrac{2}{3}\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+\dfrac{3}{2}>0\\2x-\dfrac{2}{3}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+\dfrac{3}{2}< 0\\2x-\dfrac{2}{3}>0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x>-\dfrac{3}{2}\\2x< \dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}2x< -\dfrac{3}{2}\\2x>\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< 2x< \dfrac{2}{3}\\\dfrac{2}{3}< 2x< -\dfrac{3}{2}\text{(vô lí)}\end{matrix}\right.\)
\(\Rightarrow-\dfrac{3}{4}< x< \dfrac{1}{3}\)
Vậy ...
\(\left(x-\dfrac{3}{5}\right)\left(x+\dfrac{3}{8}\right)>0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>0\\x+\dfrac{3}{8}>0\end{matrix}\right.\) 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 \(x>\dfrac{3}{5}\) \(\left(đpcm\right)\)
