Phương trình hoành độ giao điểm: \(x^2+2x+m-1=0\)
\(\Delta'=2-m>0\Rightarrow m< 2\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=m-1\end{matrix}\right.\)
\(\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]+x_1x_2=4\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(5-m\right)+m-1=4\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(5-m\right)=5-m\)
\(\Leftrightarrow x_1-x_2=1\) (\(m< 2\Rightarrow5-m\ne0\))
Kết hợp Viet ta có hệ:
\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1-x_2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-\frac{1}{2}\\x_2=-\frac{3}{2}\end{matrix}\right.\)
Mà \(m-1=x_1x_2=\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)=\frac{3}{4}\Rightarrow m=1+\frac{3}{4}=\frac{7}{4}\)
