Xét PT hoành độ giao điểm (P) và (d) có:
\(x^2=mx-m+1\Leftrightarrow x^2-mx+m-1=0\left(1\right)\)
(P) cắt (d) tại 2 điểm phân biệt\(\Leftrightarrow\) PT(1) có 2 nghiệm phân biệt
\(\Delta=\left(-m\right)^2-4\cdot\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2>0\Leftrightarrow m\ne2\)
Theo Vi-ét có:\(\left\{{}\begin{matrix}x_1+x_2=m\left(1\right)\\x_1x_2=m-1\left(2\right)\end{matrix}\right.\)
a) Theo đề bài có: \(\left|x_1\right|+\left|x_2\right|=4\)
\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=16\)
\(\Leftrightarrow x_1^2+2\left|x_1x_2\right|+x_2^2=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2\left(x_1x_2\right)+2\left|x_1x_2\right|=16\)
\(\Leftrightarrow m^2-2\left(m-1\right)+2\left|m-1\right|=16\)
TH1: \(m-1\ge0\Leftrightarrow m\ge1\)
\(\Leftrightarrow m^2-2m+2+2m-2=16\)
\(\Leftrightarrow m^2=16\Leftrightarrow m=\left[{}\begin{matrix}4\\-4\left(L\right)\end{matrix}\right.\Leftrightarrow m=4\)
TH2:\(m-1< 0\Leftrightarrow m< 1\)
\(\Leftrightarrow m^2-2m+2+2-2m=16\)
\(\Leftrightarrow m^2-4m+4=16\)
\(\Leftrightarrow\left(m-2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}m-2=4\\m-2=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=6\left(L\right)\\m=-2\end{matrix}\right.\)
\(\Leftrightarrow m=-2\)
Vậy \(m=\left\{4;-2\right\}\) là GT cần tìm
b)Theo đề bài ta có: \(x_1=9x_2\left(3\right)\)
Từ (1)(3) có:\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1=9x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x_2+x_2=m\\x_1=9x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{m}{10}\\x_1=\dfrac{9m}{10}\end{matrix}\right.\)
Thay x1, x2 vào (2) ta có:
\(\dfrac{9m}{10}\cdot\dfrac{m}{10}=m-1\)
\(\Leftrightarrow\dfrac{9m^2}{100}=m-1\)
\(\Leftrightarrow100m-100=9m^2\)
\(\Leftrightarrow9m^2-100m+100=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{10}{9}\\m=10\end{matrix}\right.\)(T/m)
Vậy \(m=\left\{10;\dfrac{10}{9}\right\}\) là GT cần tìm