Sửa đề: \(x_1\left(x_1-2x_2\right)+x_2\left(x_2-2x_1\right)=9\)
\(\Delta=\left(2m-1\right)^2-4\left(m^2+m-2\right)\)
\(=4m^2-4m+1-4m^2-4m+8=-8m+9\)
Để phương trình có hai nghiệm phân biệt thì -8m+9>0
=>-8m>-9
=>\(m<\frac98\)
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2m-1\\ x_1x_2=\frac{c}{a}=m^2+m-2\end{cases}\)
Ta có: \(x_1\left(x_1-2x_2\right)+x_2\left(x_2-2x_1\right)=9\)
=>\(x_1^2-2\cdot x_1x_2+x_2^2-2\cdot x_1x_2=9\)
=>\(x_1^2+x_2^2-4\cdot x_1x_2=9\)
=>\(\left(x_1+x_2\right)^2-6x_1x_2=9\)
=>\(\left(2m-1\right)^2-6\left(m^2+m-2\right)=9\)
=>\(4m^2-4m+1-6m^2-6m+12=9\)
=>\(-2m^2-10m+13=9\)
=>\(-2m^2-10m+4=0\)
=>\(m^2+5m-2=0\)
=>\(m^2+5m+\frac{25}{4}-\frac{33}{4}=0\)
=>\(\left(m+\frac52\right)^2=\frac{33}{4}\)
=>\(\left[\begin{array}{l}m+\frac52=\frac{\sqrt{33}}{2}\\ m+\frac52=-\frac{\sqrt{33}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}m=\frac{\sqrt{33}-5}{2}\left(nhận\right)\\ m=\frac{-\sqrt{33}-5}{2}\left(nhận\right)\end{array}\right.\)
