\(g'\left(x\right)=f'\left(\left|2x+1\right|-2\right).\left(\dfrac{1}{2\sqrt{\left(2x+1\right)^2}}\right).2\left(2x+1\right)=0\Rightarrow x=-\dfrac{1}{2}\)
TH2 : \(\left[{}\begin{matrix}\left|2x+1\right|-2=-1\\\left|2x+1\right|-2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0;x=-1\\x=3;x=-4\end{matrix}\right.\)
=> hs có 5 cực trị



