\(\Delta=m^2-2m+1+4m+4=\left(m+1\right)^2+4>0\) \(\forall m\)
Phương trình luôn có 2 nghiệm pb
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=-m-1\end{matrix}\right.\)
\(A=\left|x_1-x_2\right|\Leftrightarrow A^2=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(A^2=\left(m-1\right)^2+4m+4=m^2+2m+5\)
\(A^2=\left(m+1\right)^2+4\ge4\)
\(\Rightarrow A\ge2\) (do \(A\ge0\))
\(\Rightarrow A_{min}=2\) khi \(m+1=0\Leftrightarrow m=-1\)