a/ Pt hoành độ giao điểm
\(x^2=2(m+1)-2m-10\\↔x^2-2(m+1)+2m+10=0\)
\(Δ'=[-(m+1)]^2-1.(2m+10)=m^2+2m+1-2m-10=m^2-9=(m-3)(m+3)\)
Để (d) cắt (P) tại 2 điểm phân biệt
\(→Δ'=(m-3)(m+3)>0\\↔\left[\begin{array}{1}\begin{cases}m-3>0\\m+3>0\end{cases}\\\begin{cases}m-3<0\\m+3<0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}m>3\\m>-3\end{cases}\\\begin{cases}m<3\\m<-3\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}m>3\\m<-3\end{array}\right.\)
b/ Theo Vi-ét:
\(\begin{cases}x_1+x_2=2(m+1)\\x_1x_2=2m+10\end{cases}\)
\(A=12x_1x_2+x_1^2+x_2^2\\=12x_1x_2+(x_1+x_2)^2-2x_1x_2\\=10x_1x_2+(x_1+x_2)^2\\=10.(2m+10)+[2(m+1)]^2\\=20m+100+4(m^2+2m+1)\\=20m+100+4m^2+8m+4\\=4m^2+28m+104\\=4(m^2+7m+26)\\=4(m^2+2.\dfrac{7}{2}.m+\dfrac{49}{4})+55\\=4(m+\dfrac{7}{2})^2+55\)
Vì \( 4(m+\dfrac{7}{2})^2\ge 0→A≥55\)
\(→\min A=55\)
\(→\) Dấu "=" xảy ra khi \(m+\dfrac 7 2=0\)
\(↔m=\dfrac 7 2(TM)\)
Vậy \(\min A=55\) khi \(m=\dfrac 7 2\)
