b/ Pt có nghiệm
\(→Δ=[2(m+1)]^2-4.1.(m^2-2m+1)=4m^2+8m+4-4m^2+8m-4=16m≥0\\↔m≥0\)
c/ Theo Vi-ét:
\(\begin{cases}x_1+x_2=-2(m+1)\\x_1x_2=m^2-2m+1\end{cases}\)
\(x_1+x_1x_2+x_2+5=0\\↔-2(m+1)+m^2-2m+1+5=0\\↔m^2-2m+6-2m-2=0\\↔m^2-4m+4=0\\↔(m-2)^2=0\\↔m-2=0\\↔m=2\)
Vậy \(m=2\)