a) có 1 nghiệm \(\left[{}\begin{matrix}m+1=0;m=-1\\\left(m-1\right)^2-\left(m+1\right)\left(m-3\right)=0< =>4=0;vn\end{matrix}\right.\)
b) từ (a) luôn có 2 nghiệm mọi m khác -1
\(\left[{}\begin{matrix}x_1=\dfrac{m-3}{m+1}\\x_2=\dfrac{m+1}{m+1}\end{matrix}\right.\) \(\left\{{}\begin{matrix}x_1.x_2>0\Leftrightarrow\dfrac{m-3}{m+1}>0;m\in(-vc;-1)U\left(3;vc\right)\\x_1=2x_2\Leftrightarrow\left[{}\begin{matrix}\dfrac{m-3}{m+1}=2;-m-5=0;m=-5\\\dfrac{m-3}{m+1}=\dfrac{1}{2};m-7=0;m=7\end{matrix}\right.\end{matrix}\right.\)
kết hợp ; m =-5 ; 7