a) ta có : \(\left(x-\dfrac{1}{3}\right).\left(x+\dfrac{2}{3}\right)>0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>\dfrac{1}{3}\) hoặc \(x< \dfrac{-2}{3}\)

b) \(\left(x+\dfrac{3}{5}\right).\left(x+1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{-3}{5}\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{-3}{5}\\x>-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\-1< x< \dfrac{-3}{5}\end{matrix}\right.\) vậy \(-1< x< \dfrac{-3}{5}\)