Ta có : \(x^2-2x-2m+1=0\)
=> \(\Delta=b^{,2}-ac=1-\left(-2m+1\right)=1+2m-1=2m\)
- Để phương trình có hai nghiệm phân biệt thì :\(\Delta>0\)
=> m > 0 .
- Theo vi ét : \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=1-2m\end{matrix}\right.\)
- Ta có : \(x^2_2\left(x_1^2-1\right)+x^2_1\left(x_2^2-1\right)=8\)
=> \(\left(x_1x_2\right)^2-x^2_2+\left(x_1x_2\right)^2-x^2_1=8\)
=> \(2\left(x_1x_2\right)^2-\left(x_1^2+x_2^2\right)=8\)
=> \(2\left(x_1x_2\right)^2-\left(\left(x_1+x_2\right)^2-2x_1x_2\right)=8\)
=> \(2\left(1-2m\right)^2-\left(2^2-2\left(1-2m\right)\right)=8\)
=> \(2-8m+8m^2-4+2-4m-8=0\)
=> \(8m^2-12m-8=0\)
=> \(\left[{}\begin{matrix}m=2\left(TM\right)\\m=-\frac{1}{2}\left(L\right)\end{matrix}\right.\)
=> m = 2 .
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