\(\Leftrightarrow\left(x^2-3\right)^2+4\left|x^2-3\right|-5=0\)
Đặt \(\left|x^2-3\right|=a\ge0\)
\(\Rightarrow a^2+4a-5=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left|x^2-3\right|=1\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm2\end{matrix}\right.\)