a/ Đường thẳng (d) đi qua điểm \( (14;2020)\)
\(→2.14-m+1=2020\\↔29-m=2020\\↔m=-1991\)
b/ Pt hoành độ giao điểm
\(\dfrac{1}{2}x^2=2x-m+1\\↔\dfrac{1}{2}x^2-2x+m-1=0\\↔x^2-4x+2m-2=0\)
Để 2 đường thẳng cắt nhau tại hai điểm phân biệt
\(→Δ'=(-2)^2-1.(2m-2)>0\\↔4-2m+2>0\\↔6-2m>0\\↔2m<6\\↔m<3\)
Theo Vi-ét:
\(\begin{cases}x_1+x_2=4\\x_1x_2=2m-2\end{cases}\)
\(x_1x_2(y_1+y_2)+48=0\\↔x_1x_2(\dfrac{1}{2}x_1^2+\dfrac{1}{2}x_2^2)+48=0\\↔x_1x_2.\dfrac{1}{2}(x_1^2+x_2^2)+48=0\\↔x_1x_2.\dfrac{1}{2}(x_1^2+2x_1x_2+x_2^2-2x_1x_2)+48=0\\↔x_1x_2.\dfrac{1}{2}.[(x_1+x_2)^2-2x_1x_2]+48=0\\↔(2m-2).\dfrac{1}{2}.[4^2-2.(2m-2)]+48=0\\↔2(m-1).\dfrac{1}{2}.[16-4m+4]+48=0\\↔(m-1)(20-4m)+48=0\\↔-4m^2+24m-20+48=0\\↔-4m^2+24m+24=0\\↔m^2-6m-6=0(*)\\Δ'(*)=(-3)^2-1.(-6)=15>0\\→\left[\begin{array}{1}m_1=3+\sqrt{15}(KTM)\\m_2=3-\sqrt{15}(TM)\end{array}\right.\)
Vậy \(m=3-\sqrt{15}\)
