0=(2x+1)2
4x2 + 4x + 1 = 0
4x2 = 0 hay 4x + 1 = 0
x = 2 hay x= \(-\dfrac{1}{4}\)
(2x+1)=(2x+1)
=> (2x+1)^4 - (2x+1)^6 = 0
=> (2x+1)^4 * [1 - (2x+1)^2] = 0
=> \(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left[1-\left(2x+1\right)^2\right]=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x=0\\2x=-2\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)Vậy x\(\in\){0;-1;\(\dfrac{1}{2}\)}
