ta có : \(\left(m-1\right)\left(mx+1\right)>0\)\(\Leftrightarrow m^2x+m-mx-1>0\)
\(\Leftrightarrow m^2x-mx>1-m\) \(\Leftrightarrow x\left(m^2-m\right)>1-m\)
(*) \(m^2-m>0\Leftrightarrow m^2>m\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)
\(\Rightarrow x\left(m^2-m\right)>1-m\Leftrightarrow x>\dfrac{1-m}{m^2-m}=\dfrac{-1}{m}\)
\(\Rightarrow S=\left(\dfrac{-1}{m};+\infty\right)\)
(*) \(m^2-m< 0\Leftrightarrow m^2< m\Leftrightarrow\left\{{}\begin{matrix}m< 1\\m>-1\\me0\end{matrix}\right.\Leftrightarrow-1< m< 1+me0\)
\(\Rightarrow x\left(m^2-m\right)>1-m\Leftrightarrow x< \dfrac{1-m}{m^2-m}=\dfrac{-1}{m}\) \(\Rightarrow S=\left(-\infty;\dfrac{-1}{m}\right)\)
(*) \(m^2-m=0\Leftrightarrow\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\)
+ \(m=0\) \(\Rightarrow x\left(m^2-m\right)>1-m\Leftrightarrow0>1\left(vôlí\right)\)
+ \(m=1\)
\(\Rightarrow x\left(m^2-m\right)>1-m\Leftrightarrow0>0\left(vôlí\right)\)
\(\Rightarrow S=\varnothing\)
vậy =========-
