ĐKXĐ: \(\left\{{}\begin{matrix}\frac{-3}{2+x}\ge0\\2+x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{-3}{2+x}>0\\x\ne-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2+x< 0\\x\ne-2\end{matrix}\right.\Leftrightarrow x< -2\)
Ta có: \(\sqrt{\frac{-3}{2+x}}=2\)
\(\Leftrightarrow\sqrt{\left(\frac{-3}{2+x}\right)^2}=2^2=4\)
\(\Leftrightarrow\frac{-3}{2+x}=4\)
\(\Leftrightarrow2+x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{-3}{4}-2=\frac{-3}{4}-\frac{8}{4}=\frac{-11}{4}\)
Vậy: \(S=\left\{-\frac{11}{4}\right\}\)
ĐKXĐ : \(\left\{{}\begin{matrix}-\frac{3}{x+2}\ge0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x+2\le0\\x+2\ne0\end{matrix}\right.\)
=> \(x+2< 0\)
=> \(x< -2\)
Ta có : \(\sqrt{-\frac{3}{x+2}}=2\)
=> \(-\frac{3}{x+2}=4\)
=> \(-3=4x+8\)
=> \(x=-\frac{11}{4}\) ( TM )
Vậy ...