ĐKXĐ: \(-3\le x\le6\)
Ta có pt: \(\sqrt{3+x}+\sqrt{6-x}-\sqrt{\left(3+x\right)\left(6-x\right)}=3\)
Đặt \(a=\sqrt{3+x}+\sqrt{6-x}\left(a\ge0\right)\Rightarrow a^2=9+2\sqrt{\left(3+x\right)\left(6-x\right)}\)
\(\Rightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=\frac{a^2-9}{2}\)
Pt trở thành: \(a-\frac{a^2-9}{2}=3\Rightarrow2a-a^2+9=6\Rightarrow-a^2+2a+3=0\)
\(\Rightarrow\left(a+1\right)\left(a-3\right)=0\Rightarrow\orbr{\begin{cases}a=-1\left(l\right)\\a=3\left(n\right)\end{cases}}\)
Khi a = 3 \(\Rightarrow\sqrt{3+x}+\sqrt{6-x}=3\Rightarrow9+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)
\(\Rightarrow2\sqrt{\left(3+x\right)\left(6-x\right)}=0\Rightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=0\)
\(\Rightarrow\left(3+x\right)\left(6-x\right)=0\Rightarrow\orbr{\begin{cases}x=-3\left(n\right)\\x=6\left(n\right)\end{cases}}\)
Vậy x = -3 , x = 6