\(\sqrt{x+1}=1-2x\)
ĐKXĐ:\(-1\le x\le\dfrac{1}{2}\)
Bình phương cả hai vế ta có:
\(x+1=\left(1-2x\right)^2\Leftrightarrow x+1=4x^2-4x+1\)\(\Leftrightarrow4x^2-5x=0\Leftrightarrow4x\left(x-\dfrac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)