Đáp án:
Giải thích các bước giải:
`|x^2-1|=2x+1`
`ĐK:2x+1>=0=>x>=-1/2`
Ta có
`|x^2-1|=2x+1`
`<=>`\(\left[ \begin{array}{l}x^2-1=2x+1\\x^2-1=-2x-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^2-2x-2=0\\x^2+2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x^2-2x+1)-3=0\\x(x+2)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(x-1)^2=3\\x(x+2)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{array} \right.\\\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\ \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{array} \right.\\\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\ \end{array} \right.\)
Dựa vào điều kiện `x>=-1/2`
`=>x in {\sqrt{3}+1;0}`
Vậy `S= {\sqrt{3}+1;0}`
