Question

Answer 1

\(\Delta'=\left(m+1\right)^2-\left(4m-m^2\right)=2m^2-2m+1=\dfrac{1}{2}\left(2m-1\right)^2+\dfrac{1}{2}>0;\forall m\)

Pt luôn có 2 nghiệm pb với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=4m-m^2\end{matrix}\right.\)

\(A=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{4\left(m+1\right)^2-4\left(4m-m^2\right)}=\sqrt{8m^2-8m+4}\)

\(=\sqrt{2\left(2m-1\right)^2+2}\ge2\)

Dấu "=" xảy ra khi \(2m-1=0\Rightarrow m=\dfrac{1}{2}\)