a: \(\text{Δ}=\left[-\left(2m-1\right)\right]^2-4\cdot1\cdot\left(m^2-1\right)\)
\(=\left(2m-1\right)^2-4\left(m^2-1\right)\)
\(=4m^2-4m+1-4m^2+4=-4m+5\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-4m+5>0
=>-4m>-5
=>\(m< \dfrac{5}{4}\)
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m-1\\x_1x_2=\dfrac{c}{a}=m^2-1\end{matrix}\right.\)
\(\left(x_1-x_2\right)^2=x_1-3x_2\)
=>\(x_1-3x_2=\left(x_1+x_2\right)^2-4x_1x_2\)
=>\(x_1-3x_2=\left(2m-1\right)^2-4\left(m^2-1\right)\)
=>\(x_1-3x_2=4m^2-4m+1-4m^2+4=-4m+5\)
Do đó, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x_1-3x_2=-4m+5\\x_1+x_2=2m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-4x_2=-4m+5-2m+1=-6m+6\\x_1+x_2=2m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_2=1,5m-1,5\\x_1=2m-1-1,5m+1,5=0,5m+0,5\end{matrix}\right.\)
\(x_1x_2=m^2-1\)
=>\(\left(1,5m-1,5\right)\left(0,5m+0,5\right)=m^2-1\)
=>\(0,75\left(m^2-1\right)=m^2-1\)
=>\(m^2-1=0\)
=>\(\left[{}\begin{matrix}m=1\left(nhận\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)
