Question

`x^2 -mx-1=0` có 2 nghiệm x1, x2 thỏa mãn \(\dfrac{x^2_1+x_1-1}{x_1}-\dfrac{x_2^2+x_2-1}{x_2}=1\), tìm nghiệm

Answer 1

\(\text{Δ}=\left(-m\right)^2-4\cdot1\cdot\left(-1\right)=m^2+4>=4>0\forall m\)

=>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)

=>\(m\in\varnothing\)