a: Khi m=0 thì f(x)=-x2-x+1
f(x)<0
\(\Leftrightarrow-x^2-x+1< 0\)
\(\Leftrightarrow x^2+x-1>0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}>\dfrac{\sqrt{5}}{2}\\x+1< -\dfrac{\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{\sqrt{5}-1}{2}\\x< \dfrac{-\sqrt{5}-1}{2}\end{matrix}\right.\)
b: TH1: m=1
Pt sẽ là -2x+2=0
=>-2x=-2
hay x=1(loại)
TH2: m<>1
\(\text{Δ}=\left(m+1\right)^2-4\left(m-1\right)\left(m+1\right)\)
\(=m^2+2m+1-4m^2+4=-3m^2+2m+5\)
Để f(x) vô nghiệm thì \(3m^2-2m-5>0\)
\(\Leftrightarrow\left(3m-5\right)\left(m+1\right)>0\)
=>m>5/3 hoặc m<-1
