ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge-3\end{matrix}\right.\) \(\Rightarrow x+y+1\ge0\)
Bình phương 2 vế giả thiết:
\(\left(x+y+1\right)^2=4\left(x+y+1+2\sqrt{\left(x-2\right)\left(y+3\right)}\right)\)
\(\Rightarrow\left(x+y+1\right)^2\le4\left(x+y+1+x+y+1\right)=5\left(x+y+1\right)\)
\(\Rightarrow x+y+1\le5\Rightarrow x+y\le4\)
Mặt khác:
\(\left(x+y+1\right)^2=4\left(x+y+1+2\sqrt{\left(x-2\right)\left(y+3\right)}\right)\ge4\left(x+y+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+1\ge4\\x+y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y\ge3\\x+y=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}3\le S\le4\\S=-1\end{matrix}\right.\)
