1: Thay y=-8 vào (P), ta được:
\(-2x^2=-8\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
vậy: Các điểm thỏa mãn là A(2;-4); B(-2;-4)
b: Phương trình hoành độ giao điểm là:
\(-2x^2=-2\left(m+1\right)x+\dfrac{1}{2}m^2-1\)
=>\(-2x^2+2\left(m+1\right)x-\dfrac{1}{2}m^2+1=0\)
\(\text{Δ}=\left(2m+2\right)^2-4\cdot\left(-2\right)\cdot\left(-\dfrac{1}{2}m^2+1\right)\)
\(=4m^2+8m+4+8\left(-\dfrac{1}{2}m^2+1\right)\)
\(=4m^2+8m+4-4m^2+8=8m+12\)
Để (d) cắt (P) tại hai điểm phân biệt thì 8m+12>0
=>8m>-12
=>\(m>-\dfrac{3}{2}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m+1\right)}{-2}=m+1\\x_1x_2=\dfrac{c}{a}=\dfrac{-\dfrac{1}{2}m^2+1}{-2}=\dfrac{\dfrac{1}{2}m^2-1}{2}=\dfrac{1}{4}m^2-\dfrac{1}{2}\end{matrix}\right.\)
\(x_1^2+\left(m+1\right)x_2=2x_1x_2+\dfrac{15}{2}\)
=>\(x_1^2+x_2\left(x_1+x_2\right)=2\cdot\left(\dfrac{1}{4}m^2-\dfrac{1}{2}\right)+\dfrac{15}{2}\)
=>\(\left(x_1^2+x_1^2\right)+x_1x_2=\dfrac{1}{2}m^2+\dfrac{13}{2}\)
=>\(\left(x_1+x_2\right)^2-x_1x_2=\dfrac{1}{2}m^2+\dfrac{13}{2}\)
=>\(\left(m+1\right)^2-\left(\dfrac{1}{4}m^2-\dfrac{1}{2}\right)=\dfrac{1}{2}m^2+\dfrac{13}{2}\)
=>\(m^2+2m+1-\dfrac{1}{4}m^2+\dfrac{1}{2}-\dfrac{1}{2}m^2-\dfrac{13}{2}=0\)
=>\(\dfrac{1}{4}m^2+2m-5=0\)
=>\(m^2+8m-20=0\)
=>(m+10)(m-2)=0
=>\(\left[{}\begin{matrix}m=-10\left(loại\right)\\m=2\left(nhận\right)\end{matrix}\right.\)
