ĐKXĐ \(2x+1\ge0\Rightarrow x\ge-\frac{1}{2}\)
Khi đó |x2 - 1| = 2x + 1
,=> \(\orbr{\begin{cases}x^2-1=2x+1\\x^2-1=-2x-1\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-2x=2\\x^2+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-2x+1=3\\x^2+2x+1=1\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=3\\\left(x+1\right)^2=1\end{cases}}\)
Nếu (x - 2)2 = 3
=> \(\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{cases}}\)(tm)
Nếu (x + 1)2 = 1
=> \(\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\left(\text{loại}\right)\end{cases}}\Rightarrow x=0\)
