\(TH1:m=1.\\ \Rightarrow0-2\left(1+3\right)x-1+2=0.\\ \Leftrightarrow-8x+1=0.\\ \Leftrightarrow x=\dfrac{1}{8}.\)
\(\Rightarrow\) \(m=1\) thỏa mãn yêu cầu bài.
\(TH2:m\ne1.\\ \Delta'=m^2+6m+9+\left(m-1\right)\left(m-2\right).\\ =m^2+6m+9-m^2-2m-m+2.\\ =3m+11\ge0.\\ \Leftrightarrow m\ge\dfrac{-11}{3}.\)
\(\Rightarrow m\ge\dfrac{-11}{3};m\ne1.\)