Pt hoành độ giao điểm:
\(-\dfrac{1}{2}x^2=2x-m^2-1\Leftrightarrow x^2+4x-2\left(m^2+1\right)=0\)
\(ac=-2\left(m^2+1\right)< 0\) ; \(\forall m\Rightarrow\) (d) luôn cắt (P) tại 2 điểm có hoành độ trái dấu
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-4\\x_1x_2=-2\left(m^2+1\right)\end{matrix}\right.\)
Ta có: \(\dfrac{1}{x_1}=\dfrac{1}{\left|x_2\right|}+\dfrac{1}{2}>0\Rightarrow x_1>0\Rightarrow x_2< 0\Rightarrow\dfrac{1}{\left|x_2\right|}=-\dfrac{1}{x_2}\)
Do đó:
\(\dfrac{1}{x_1}=\dfrac{1}{\left|x_2\right|}+\dfrac{1}{2}\Leftrightarrow\dfrac{1}{x_1}=-\dfrac{1}{x_2}+\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{2}\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{-4}{-2\left(m^2+1\right)}=\dfrac{1}{2}\Leftrightarrow m^2+1=4\)
\(\Leftrightarrow m^2=3\Rightarrow m=\pm\sqrt{3}\)
