ĐK: \(-3\le x\le6\)
\(\sqrt{x+3}+\sqrt{6-x}-\sqrt{\left(x-3\right)\left(6-x\right)}=3\)(1)
Đặt a=\(\sqrt{x+3}\left(a\ge0\right)\),b=\(\sqrt{6-x}\left(b\ge0\right)\)\(\Leftrightarrow a^2+b^2=9\)
Vậy (1)\(\Leftrightarrow a+b-ab=3\)
Vậy ta có hệ phương trình \(\left\{{}\begin{matrix}a^2+b^2=9\\a+b-ab=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=9\\a+b-ab=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=9\\2\left(a+b\right)-2ab=6\end{matrix}\right.\)\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)-15=0\Leftrightarrow\left(a+b-3\right)\left(a+b+5\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a+b-3=0\\a+b+5=0\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a+b=3\)
Vậy \(\sqrt{x+3}+\sqrt{6-x}=3\)
Mà \(\sqrt{x+3}+\sqrt{6-x}-\sqrt{\left(x+3\right)\left(6-x\right)}=3\)
Suy ra \(\sqrt{\left(x+3\right)\left(6-x\right)}=0\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\6-x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
Vậy S={-3;6}
