Question

[kurumi cute]

Làm câu vi et thôi

 

Answer 1

[Nguyễn Lê Phước Thịnh]

2:

\(\text{Δ}=\left[-\left(2m-1\right)\right]^2-4\cdot1\cdot\left(m^2-m-2\right)\)

\(=4m^2-4m+1-4m^2+4m+8=9>0\)

=>Phương trình luôn có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{2m-1-\sqrt{9}}{2}=\dfrac{2m-1-3}{2}=m-2\\x=\dfrac{2m-1+3}{2}=\dfrac{2m+2}{2}=m+1\end{matrix}\right.\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m-1\\x_1x_2=\dfrac{c}{a}=m^2-m-2\end{matrix}\right.\)

\(\dfrac{x_1^4+x_1^3+m^2-m-2}{x_1}-\dfrac{x_2^4+x_2^3+m^2-m-2}{x_2}=-7m^2+4m+24\)

=>\(x_1^3+x_1^2+\dfrac{x_1x_2}{x_1}-x_2^3-x_2^2-\dfrac{x_1x_2}{x_2}=-7m^2+4m+24\)

=>\(\left(x_1^3-x_2^3\right)+\left(x_1^2-x_2^2\right)+\left(x_2-x_1\right)=-7m^2+4m+24\)

=>\(\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)+\left(x_1-x_2\right)\left(x_1+x_2\right)-\left(x_1-x_2\right)=-7m^2+4m+24\)

=>)\(\left(x_1-x_2\right)\left(x_1^2+x_2x_1+x_2^2+x_1+x_2-1\right)=-7m^2+4m+24\)(1)

TH1: \(x_1=m-2;x_2=m+1\)

(1) sẽ tương đương với:

\(\left(m-2-m-1\right)\left[\left(m-2\right)^2+\left(m-2\right)\left(m+1\right)+\left(m+1\right)^2+m-2+m+1-1\right]=-7m^2+4m+24\)

=>\(-3\left[m^2-4m+4+m^2-m-2+m^2+2m+1+2m-2\right]=-7m^2+4m+24\)

=>\(-3\left(3m^2-m+1\right)+7m^2-4m-24=0\)

=>\(-9m^2+3m-3+7m^2-4m-24=0\)

=>\(-2m^2-m-27=0\)

=>\(m\in\varnothing\)

TH2: \(x_1=m+1;x_2=m-2\)

(1) sẽ trở thành:

\(\left(m+1-m+2\right)\left[\left(m+1\right)^2+\left(m+1\right)\left(m-2\right)+\left(m-2\right)^2+2m-1-1\right]=-7m^2+4m+24\)

=>\(3\left(m^2+2m+1+m^2-m-2+m^2-4m+4+2m-2\right)=-7m^2+4m+24\)

=>\(3\left(3m^2-m+1\right)+7m^2-4m-24=0\)

=>\(9m^2-3m+3+7m^2-4m-24=0\)

=>\(16m^2-7m-21=0\)

=>\(m=\dfrac{7\pm\sqrt{1393}}{32}\)