\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\)
\(\Rightarrow x>2\) hoặc \(x< \dfrac{-2}{3}\)
Vậy \(\left[{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\)
\(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\Rightarrow x-2\text{ và }x+\dfrac{2}{3}\text{ cùng dấu}\\ \left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\Rightarrow x>2\\ \left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\Rightarrow x< \dfrac{-2}{3}\\ \text{Vậy }\left\{{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\)
