\(\Delta'=1^2-\left(2m-10\right)=1-2m+10=11-2m\)
Để phương trình có nghiệm
\(\Rightarrow\Delta\ge0\)
\(\Rightarrow11-2m\ge0\Rightarrow2m\le11\Rightarrow m\le\frac{11}{2}\)
Theo hệ thức Vi-ét
\(\left\{{}\begin{matrix}x_1+x_2=-2\left(m+1\right)\\x_1x_2=2m-10\end{matrix}\right.\)
Ta có
\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{x_1^2+x_2^2}{x_1x_2}=\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\frac{\left[2\left(m+1\right)\right]^2-2\left(2m-10\right)}{2m-10}\)
\(=\frac{2m^2+4m+2-4m+2}{2m-10}=\frac{2m^2+4}{2m-10}\)
Mà \(\frac{x_1}{x_2}+\frac{x_2}{x_1}=2\)
\(\Rightarrow2m^2+4=2\left(2m+10\right)\Leftrightarrow2m^2+4-4m-20=0\Leftrightarrow2m^2-4m-16=0\)
\(2\left(m^2-2m+1\right)=18\)
\(\Rightarrow2\left(m-1\right)^2=18\Rightarrow\left(m-1\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}m-1=3\\m-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=4\\m=-2\end{matrix}\right.\) ( thỏa mãn \(m\le\frac{11}{2}\) )
Vậy m \(\in\left\{-2;4\right\}\)
