\(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{2}{3}\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{2}{3}\end{matrix}\right.\)
=> \(-\dfrac{2}{3}< x< 3\)
p/s : Không biết đúng hay sai bởi vì đang áp dụng cách mới :v
\(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)
TH1:
\(\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow x>3\)
TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{2}{3}\)
Vậy để \(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)
thì \(x>3\) hoặc \(x< -\dfrac{2}{3}\)
