\(\text{Δ}=\left(3m-1\right)^2-4\cdot1\cdot\left(2m^2-m\right)\)
\(=9m^2-6m+1-8m^2+4m\)
\(=m^2-2m+1=\left(m-1\right)^2\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>\(\left(m-1\right)^2>0\)
=>\(m-1\ne0\)
=>\(m\ne1\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3m-1\\x_1x_2=\dfrac{c}{a}=2m^2-m\end{matrix}\right.\)
\(\left|x_1-x_2\right|-2=0\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=2\)
=>\(\sqrt{\left(3m-1\right)^2-4\left(2m^2-m\right)}=2\)
=>\(\sqrt{9m^2-6m+1-8m^2+4m}=2\)
=>\(\sqrt{m^2-2m+1}=2\)
=>\(\sqrt{\left(m-1\right)^2}=2\)
=>|m-1|=2
=>\(\left[{}\begin{matrix}m-1=2\\m-1=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=3\\m=1\end{matrix}\right.\)
