Ta có: \(\dfrac{\left(x-2\right)\left(x+3\right)}{x-1}\)>0
Th1:\(\left\{{}\begin{matrix}\left(x-2\right)\left(x+3\right)>0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2>0\\x+3>0\\x-1>0\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+3< 0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>2\\x>-3\\x>1\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x< 2\\x< -3\\x>1\end{matrix}\right.\)( Vô lý )
=>x>2
Th2:\(\left\{{}\begin{matrix}\left(x-2\right)\left(x+3\right)< 0\\x-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2>0\\x+3< 0\\x-1< 0\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+3>0\\x-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>2\\x< -3\\x< 1\end{matrix}\right.\) ( Vô lý ) Hoặc \(\left\{{}\begin{matrix}x< 2\\x>-3\\x< 1\end{matrix}\right.\)
=> -3<x<1
Vậy nghiệm của phương trình là
x>2 hoặc -3<x<1
