\(\Delta=\left(2m+1\right)^2-4\left(2m-4\right)=\left(2m-1\right)^2+16>0;\forall m\)
Phương trình luôn có 2 nghiệm pb
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m+1\\x_1x_2=2m-4\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=5\)
\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=25\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|-25=0\)
\(\Leftrightarrow\left(2m+1\right)^2-2\left(2m-4\right)+2\left|2m-4\right|-25=0\)
- Với \(m\ge2\)
\(\Leftrightarrow\left(2m+1\right)^2-25=0\Rightarrow\left[{}\begin{matrix}2m+1=5\\2m+1=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=2\\m=-3\left(l\right)\end{matrix}\right.\)
- Với \(m< 2\)
\(\Leftrightarrow\left(2m+1\right)^2-4\left(2m-4\right)-25=0\)
\(\Leftrightarrow4m^2-4m-8=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=2\left(l\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=-1\\m=2\end{matrix}\right.\)
