ĐKXĐ: ...
Đặt \(\sqrt{x^2+3}+\sqrt{6-x^2}=a>0\Rightarrow\sqrt{18+3x^2-x^4}=\frac{a^2-9}{2}\)
\(\Leftrightarrow a=3+\frac{a^2-9}{2}\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x^2+3}+\sqrt{6-x^2}=3\)
\(\Leftrightarrow9+2\sqrt{\left(x^2+3\right)\left(6-x^2\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x^2+3\right)\left(6-x^2\right)}=0\)
\(\Leftrightarrow6-x^2=0\Rightarrow x=\pm\sqrt{6}\)