ĐK: \(-4\le x\le1\)
<=>\(\sqrt{1-x}=3-\sqrt{4+x}\)
<=>\(1-x=9+4+x-6\sqrt{4+x}\)
<=>\(0=12+2x-6\sqrt{4+x}\)
<=>\(6+x-3\sqrt{4+x}=0\)
<=>\(6+x=3\sqrt{4+x}\)
<=>\(36+12x+x^2=9\left(4+x\right)\)
<=>\(36+12x+x^2=36+9x\)
<=>\(x^2+3x=0\)
<=>\(x\left(x+3\right)=0\)
<=>\(\left\{{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)(nhận)
Vậy S={-3;0}