Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x^2=mx-\dfrac{1}{2}m^2+m+1\)
=>\(\dfrac{1}{2}x^2-mx+\dfrac{1}{2}m^2-m-1=0\)
\(\text{Δ}=\left(-m\right)^2-4\cdot\dfrac{1}{2}\left(\dfrac{1}{2}m^2-m-1\right)\)
\(=m^2-2\left(\dfrac{1}{2}m^2-m-1\right)\)
\(=m^2-m^2+2m+2=2m+2\)
Để (P) cắt (d) tại hai điểm phân biệt thì Δ>0
=>2m+2>0
=>2m>-2
=>m>-1
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{m}{\dfrac{1}{2}}=2m\\x_1x_2=\dfrac{c}{a}=\dfrac{\dfrac{1}{2}m^2-m-1}{\dfrac{1}{2}}=2\left(\dfrac{1}{2}m^2-m-1\right)=m^2-2m-2\end{matrix}\right.\)
\(\left|x_1-x_2\right|=2\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=2\)
=>\(\sqrt{\left(2m\right)^2-4\left(m^2-2m-2\right)}=2\)
=>\(\sqrt{4m^2-4m^2+8m+8}=2\)
=>8m+8=4
=>8m=-4
=>\(m=-\dfrac{1}{2}\)(nhận)
2 =3