\(\text{Δ}=\left(-m\right)^2-4\cdot1\cdot\left(-1\right)=m^2+4>=4>0\forall m\)
=>Phương trình luôn có hai nghiệm x1,x2
Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=m;x_1x_2=\dfrac{c}{a}=-1\)
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=m^2+4\)
=>\(x_1-x_2=\sqrt{m^2+4}\)
\(\dfrac{x_1^2+x_1-1}{x_1}-\dfrac{x_2^2+x_2-1}{x_2}=1\)
=>\(x_1+1-\dfrac{1}{x_1}-x_2-1+\dfrac{1}{x_2}=1\)
=>\(\left(x_1-x_2\right)-\left(\dfrac{1}{x_1}-\dfrac{1}{x_2}\right)=1\)
=>\(\left(x_1-x_2\right)+\dfrac{x_1-x_2}{x_1x_2}=1\)
=>>\(\left(x_1-x_2\right)-\left(x_1-x_2\right)=1\)
=>0=1(vô lý)
=>\(m\in\varnothing\)