\(\Delta'=\left(m+1\right)^2-\left(4m-m^2\right)=\dfrac{1}{2}\left(2m-1\right)^2+\dfrac{1}{2}>0;\forall m\)
Phương trình luôn có 2 nghiệm phân biệt
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=4m-m^2\end{matrix}\right.\)
\(A=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(A=\sqrt{4\left(m+1\right)^2-4\left(4m-m^2\right)}=\sqrt{8m^2-8m+4}\)
\(A=\sqrt{2\left(2m-1\right)^2+2}\ge\sqrt{2}\)
\(A_{min}=\sqrt{2}\) khi \(2m-1=0\Rightarrow m=\dfrac{1}{2}\)
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