Theo bài ra ta có: (x-5).(x+3)>0
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>5\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 5\\x< -3\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5< x>-3\\5>x< -3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>5\\x< -3\end{matrix}\right.\)
Vậy x\(\in\varnothing\)
Giải:
\(\left(x-5\right)\left(x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>5\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 5\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5< x>-3\\5>x< -3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< -3\end{matrix}\right.\)
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