Ta có: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4m^2\)
\(=4\left(m-1\right)^2-4m^2\)
\(=4\left(m^2-2m+1\right)-4m^2\)
\(=4m^2-8m+4-4m^2=-8m+4\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-8m+4>0
=>-8m>-4
=>\(m< \dfrac{4}{8}=\dfrac{1}{2}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m^2\end{matrix}\right.\)
\(\dfrac{x_1^2}{x_2}+\dfrac{x_2^2}{x_1}=-5\left(x_1+x_2\right)\)
=>\(\dfrac{x_1^3+x_2^3}{x_1x_2}=-5\left(x_1+x_2\right)\)
=>\(\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=-5\cdot x_1x_2\cdot\left(x_1+x_2\right)\)
=>\(\left(x_1+x_2\right)^3+2\cdot x_1x_2\left(x_1+x_2\right)=0\)
=>\(\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2+2x_1x_2\right]=0\)
=>\(\left(2m-2\right)\left[\left(2m-2\right)^2+2\cdot m^2\right]=0\)
=>\(2\left(m-1\right)\left[4m^2-8m+4+2m^2\right]=0\)
=>\(\left(m-1\right)\left(3m^2-4m+2\right)=0\)
=>\(\left[{}\begin{matrix}m-1=0\\3m^2-4m+2=0\end{matrix}\right.\Leftrightarrow m=1\)(loại)
=>\(m\in\varnothing\)
