Ta có:
\(\Delta^'=\left[-\left(2m+1\right)\right]^2-\left(4m^2+4m-3\right)\)
\(=4m^2+4m+1-4m^2-4m+3=4>0\left(\forall m\right)\)
=> PT luôn có 2 nghiệm phân biệt:
\(\hept{\begin{cases}x_1=\frac{2m+1-2}{1}=2m-1\\x_2=\frac{2m+1+2}{1}=2m+3\end{cases}}\) vì \(x_1< x_2\)
Ta có: \(\left|x_1\right|=2\left|x_2\right|\Leftrightarrow\left|2m-1\right|=2\left|2m+3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2m-1=2\left(2m+3\right)\\1-2m=2\left(2m+3\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}2m=-7\\6m=-5\end{cases}}\Rightarrow\orbr{\begin{cases}m=-\frac{7}{2}\\m=-\frac{5}{6}\end{cases}}\)
Vậy \(m\in\left\{-\frac{7}{2};-\frac{5}{6}\right\}\)