Áp dụng phương trình hoành độ giao điểm
\(x^2-\left(m-2\right)x-m+7=-x+1\)
\(\Leftrightarrow x^2+\left(3-m\right)x+6-m=0\)
\(\Delta=\left(3-m\right)^2-4\left(6-m\right)\)
\(\Delta=m^2-2m-15\)
ĐK:\(\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)
\(x_1=\frac{m-3+\sqrt{m^2-2m-15}}{2}\)
\(x_2=\frac{m-3-\sqrt{m^2-2m-15}}{2}\)
\(y_1=\frac{m-3+\sqrt{m^2-2m-15}}{-2}+1\)
\(y_2=\frac{m-3-\sqrt{m^2-2m-15}}{-2}+1\)
Theo đề bài :
\(\left(y_1+y_2\right)^2=12\)
\(\Leftrightarrow\left(\frac{2m-6}{-2}+2\right)^2=12\)
\(\Leftrightarrow\left(5-m\right)^2=12\)
\(\Leftrightarrow m^2-10m+13=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=5+2\sqrt{3}\left(nhan\right)\\m=5-2\sqrt{3}\left(loai\right)\end{matrix}\right.\)
