ĐK: \(\left\{{}\begin{matrix}x\ge-3\\x\le6\end{matrix}\right.\)
Đặt \(t=\sqrt{x+3}+\sqrt{6-x}\left(t\ge0\right)\Leftrightarrow t^2=9+2\sqrt{\left(x+3\right)\left(6-x\right)}\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=\dfrac{t^2-9}{2}\)
Thay vào pt:
\(\Leftrightarrow t-\dfrac{t^2-9}{2}=3\\ \Leftrightarrow2t-t^2+9=6\\ \Leftrightarrow t^2-2t-3=0\Leftrightarrow\left(t+1\right)\left(t-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=-1\left(ktm\right)\\t=3\left(tm\right)\end{matrix}\right.\Leftrightarrow t=3\\ \Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=\dfrac{9-9}{2}=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
