Phương trình hoành độ giao điểm là:
\(\left(2m-1\right)x^2=2\left(m+4\right)x-5m-2\)
=>\(\left(2m-1\right)x^2-\left(2m+8\right)x+5m+2=0\)
Để (P) cắt (d) tại hai điểm phân biệt thì
\(\left\{{}\begin{matrix}2m-1\ne0\\\text{Δ}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\\left(2m+8\right)^2-4\left(2m-1\right)\left(5m+2\right)>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\4m^2+32m+64-4\left(10m^2+4m-5m-2\right)>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\4m^2+32m+64-40m^2+4m+8>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\-36m^2+36m+72>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\m^2-m-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\\left(m-2\right)\left(m+1\right)< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\-1< m< 2\end{matrix}\right.\)
Theo vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-\left(-2m-8\right)}{2m-1}=\dfrac{2m+8}{2m-1}\\x_1x_2=\dfrac{5m+2}{2m-1}\end{matrix}\right.\)
\(x_1^2+x^2_2=2x_1x_2+16\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2-2x_1x_2=16\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=16\)
=>\(\left(\dfrac{2m+8}{2m-1}\right)^2-4\cdot\dfrac{5m+2}{2m-1}=16\)
=>\(\dfrac{\left(2m+8\right)^2-4\left(5m+2\right)\left(2m-1\right)}{\left(2m-1\right)^2}=16\)
=>\(\dfrac{4m^2+32m+64-4\left(10m^2-m-2\right)}{\left(2m-1\right)^2}=16\)
=>\(-36m^2+36m+72=16\left(4m^2-4m+1\right)\)
=>\(-36m^2+36m+72=64m^2-64m+16\)
=>\(-100m^2+100m+56=0\)
=>\(\left[{}\begin{matrix}m=\dfrac{7}{5}\left(nhận\right)\\m=-\dfrac{2}{5}\left(nhận\right)\end{matrix}\right.\)
