Giải:
a) \(\left(2x+4\right)\left(x-3\right)>0\)
* TH1:
\(\left\{{}\begin{matrix}2x+4>0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>3\end{matrix}\right.\Leftrightarrow x>3\)
* TH2:
\(\left\{{}\begin{matrix}2x+4< 0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< 3\end{matrix}\right.\Leftrightarrow x< 2\)
Vậy \(x>3\) hoặc \(x< 2\).
b) \(\dfrac{x+5}{x-1}< 0\)
* TH1:
\(\left\{{}\begin{matrix}x+5>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-5\\x< 1\end{matrix}\right.\Leftrightarrow-5< x< 1\)
\(\Leftrightarrow x\in\left\{-4;-3;-2;-1;0\right\}\)
* TH2:
\(\left\{{}\begin{matrix}x+5< 0\\x-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -5\\x>1\end{matrix}\right.\Leftrightarrow-5>x>1\)
\(\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}\).
c) \(\left(x-2\right)\left(x+5\right)< 0\)
* TH1:
\(\left\{{}\begin{matrix}x-2>0\\x+5< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -5\end{matrix}\right.\Leftrightarrow2< x< -5\)
\(\Leftrightarrow x\in\left\{\varnothing\right\}\)
* TH2:
\(\left\{{}\begin{matrix}x-2< 0\\x+5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-5\end{matrix}\right.\Leftrightarrow2>x>-5\)
\(\Leftrightarrow x\in\left\{-4;-3;-2;-1;0;1\right\}\)
Vậy \(x\in\left\{-4;-3;-2;-1;0;1\right\}\).
Chúc bạn học tốt!
