a: Thay m=-2 vào (d), ta được:
y=2x-2-1=2x-3
Phương trình hoành độ giao điểm là:
\(-x^2=2x-3\)
=>\(x^2+2x-3=0\)
=>(x+3)(x-1)=0
=>\(\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
Khi x=-3 thì \(y=-x^2=-\left(-3\right)^2=-9\)
Khi x=1 thì \(y=-x^2=-1^2=-1\)
Vậy: (d) giao (P) tại A(-3;-9); B(1;-1)
b: Phương trình hoành độ giao điểm là:
\(-x^2=2x+m-1\)
=>\(x^2+2x+m-1=0\)
\(\text{Δ}=2^2-4\left(m-1\right)\)
=4-4m+4=-4m+8
Để (d) cắt (P) tại hai điểm phân biệt thì Δ>0
=>-4m+8>0
=>-4m>-8
=>m<2
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(=\left(-2\right)^2-4\left(m-1\right)\)
=4-4m+4=-4m+8
=>\(x_1-x_2=\pm\sqrt{-4m+8}\)
\(x_1^3-x_2^3+x_1x_2=4\)
=>\(\left(x_1-x_2\right)^3+3x_1x_2\left(x_1-x_2\right)+x_1x_2=4\)
=>\(\left(x_1-x_2\right)\left[\left(x_1-x_2\right)^2+3x_1x_2\right]+x_1x_2=4\)
=>\(\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]+x_1x_2=4\)
=>\(\left(x_1-x_2\right)\left[\left(-2\right)^2-\left(m-1\right)\right]+m-1=4\)
=>\(\left(x_1-x_2\right)\left(4-m+1\right)+m-5=0\)
=>\(\left(x_1-x_2\right)\left(5-m\right)-\left(5-m\right)=0\)
=>\(\left(5-m\right)\left(x_1-x_2-1\right)=0\)
=>\(\left[{}\begin{matrix}5-m=0\\x_1-x_2=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=5\\-\sqrt{-4m+8}=1\left(loại\right)\\\sqrt{-4m+8}=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=5\\-4m+8=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=5\\-4m=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=5\left(loại\right)\\m=\dfrac{7}{4}\left(nhận\right)\end{matrix}\right.\)
