TH1: m=2
Pt trở thành \(-2\left(2+1\right)x+2\cdot2-6=0\)
=>-6x-2=0
hay x=-1/3(loại)
TH2: m<>2
\(\text{Δ}=\left(2m+2\right)^2-4\left(m-2\right)\left(2m-6\right)\)
\(=4m^2+8m+4-\left(4m-8\right)\left(2m-6\right)\)
\(=4m^2+8m+4-8m^2+24m+16m-48\)
\(=-4m^2+48m-44\)
\(=-4\left(m^2-12m+11\right)\)
\(=-4\left(m-1\right)\left(m-11\right)\)
Để phương trình có hai nghiệm dương phân biệt thì:
\(\left\{{}\begin{matrix}\left(m-1\right)\left(m-11\right)< 0\\\dfrac{2\left(m+1\right)}{m-2}>0\\\dfrac{2m-6}{m-2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1< m< 11\\m\in\left(-\infty;-1\right)\cup\left(2;+\infty\right)\\m\in\left(-\infty;2\right)\cup\left(3;+\infty\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1< m< 11\\m\in\left(-\infty;-1\right)\cup\left(3;+\infty\right)\end{matrix}\right.\Leftrightarrow m\in\left(3;11\right)\)
