\(f\left(x\right)=\left(2-m\right)x^2+2\left(m-3\right)x+1-m\)
\(f\left(x\right)>0\) vô nghiệm => \(f\left(x\right)\le0\) có nghiệm ∀ x∈ R
th1 a=0
<=> m=2
vs m = 2 thế vào pt f(x) ta đc
\(\left(2-2\right)x^2+2\left(2-3\right)x+1-2\le0\)
<=> \(-6x+3\le0\)
\(< =>x\ge\dfrac{3}{6}\)
=> m=2 loại
th2 a ≠ 0
với mọi x thuộc R
\(\left\{{}\begin{matrix}a< 0\\\Delta\le0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}2-m< 0\\\left(m-3\right)^2-\left(2-m\right)\left(1-m\right)\le0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}m>2\\m^2-6m+9-\left(2-2m-m+m^2\right)\le0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}m>2\\-3m+7\le0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}m>2\\m\ge\dfrac{7}{3}\end{matrix}\right.\)
=>\(S=\text{[}\dfrac{7}{3};+\infty\text{)}\)
