\(g'\left(x\right)=f'\left(\left|2x+1\right|+2m-2020\right)\left(\dfrac{1}{2\sqrt{\left(2x+1\right)^2}}\right).2\left(2x+1\right)=0\)
TH1 : \(\Rightarrow x=-\dfrac{1}{2}\)
TH2 : \(f'\left(\left|2x+1\right|+2m-2020\right)=0\Rightarrow\left[{}\begin{matrix}\left|2x+1\right|+2m-2020=-2\\\left|2x+1\right|+2m-2020=0\\\left|2x+1\right|+2m-2020=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2m-2020=-2-\left|2x+1\right|\\2m-2020=-\left|2x+1\right|\\2m-2020=1-\left|2x+1\right|\end{matrix}\right.\)
\(u\left(x\right)=-\left|2x+1\right|\Rightarrow u'\left(x\right)=-\dfrac{2}{2\sqrt{2x+1}}\left(2x+1\right)=0\)
Để hs có 7 cực trị, do x =-1/2 => hs cần 6 cực trị
=> 2m - 2020 < -2 <=> 2m < 2018 <=> m < 1009
mà m dương => 0 < m < 1009 => 1008 giá trị => B
