$b)(m-2)x^2+(m-2)x+m \le 0(*)\\ +)m=2(*)\Leftrightarrow 2 \le 0 $
$\Rightarrow$ Bất phương trình vô nghiệm
$+)m\ne2,$ cần tìm $m$ để $(m-2)x^2+(m-2)x+m>0 \, \forall\,x$
$\Leftrightarrow\left\{\begin{array}{l} \Delta <0\\m-2>0\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} (m-2)^2-4(m-2)m <0\\m-2>0\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} m^2-4m+4-4m^2+8m <0\\m-2>0\end{array} \right. \\ \Leftrightarrow\left\{\begin{array}{l} -3m^2+4m+4 <0\\m-2>0\end{array} \right. \\ \Leftrightarrow\left\{\begin{array}{l} \left[\begin{array}{l} m>2\\m<-\dfrac{2}{3}\end{array} \right.\\ \\m>2\end{array} \right. \\ \Leftrightarrow m>2$
Kết hợp TH1
$m\ge2 thì (*)$ vô nghiệm
$c) (m+1)x^2-2(m-1)x+3m-3\le0(**)\\ +)m=-1(**)\Leftrightarrow 4x-6 \le 0 \Leftrightarrow x\le1,5$
$+)m\ne1,$ cần tìm $m$ để $(m+1)x^2-2(m-1)x+3m-3>0 \, \forall\,x$
$\Leftrightarrow\left\{\begin{array}{l} \Delta' <0\\m+1>0\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} (m-1)^2-(m+1)(3m-3) <0\\m>-1\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} m^2-2m+1-3m^2+3 <0\\m>-1\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} -2m^2-2m+4 <0\\m>-1\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} \left[\begin{array}{l} m>1\\m<-2\end{array} \right.\\m>-1\end{array} \right.\\ m>1$
