Xét PT hoành độ giao điểm (P) và (d) có:
\(x^2=mx-m+1\Leftrightarrow x^2-mx+m-1=0\) (1)
(P) cắt (d) tại 2 điểm phân biệt \(\Leftrightarrow\) PT(1) có 2 nghiệm phân biệt
\(\Delta=\left(-m\right)^2-4\cdot1\cdot\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2>0\Leftrightarrow m\ne2\)
Theo Vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1\cdot x_2=m-1\end{matrix}\right.\)
Theo đề bài ta có: \(\left|x_1\right|+\left|x_2\right|=4\)
\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=16\)
\(\Leftrightarrow x_1^2+2\left|x_1x_2\right|+x_2^2=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow m^2-2\left(m-1\right)+2\left|m-1\right|=16\left(2\right)\)
TH1: \(m-1\ge0\Leftrightarrow m\ge1\)
\(\Leftrightarrow m^2-2\left(m-1\right)+2\left(m-1\right)=16\)
\(\Leftrightarrow m^2=16\)
\(\Leftrightarrow m=\left[{}\begin{matrix}4\\-4\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow m=4\)
TH2: \(m-1< 0\Leftrightarrow m< 1\)
\(\Leftrightarrow m^2-2\left(m-1\right)+2\left(1-m\right)=16\)
\(\Leftrightarrow m^2-2m+2+2-2m=16\)
\(\Leftrightarrow m^2-4m+4=16\)
\(\Leftrightarrow\left(m-2\right)^2=16\)
\(\Leftrightarrow m-2=\pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}m=6\left(L\right)\\m=-2\end{matrix}\right.\)
\(\Leftrightarrow m=-2\)
2 =3
