\(5x^4-2x^2-3x^2\sqrt{x^2+2}=4\)
Đặt \(\sqrt{x^2+2}=t>0\Rightarrow x^2=t^2-2\)
\(5\left(t^2-2\right)^2-2\left(t^2-2\right)-3t\left(t^2-2\right)-4=0\)
\(\Leftrightarrow5t^4-3t^3-22t^2+6t+20=0\)
Nhận thấy \(t=0\) không phải nghiệm, chia 2 vế cho \(t^2\)
\(\Rightarrow5\left(t^2+\frac{4}{t^2}\right)-3\left(t-\frac{2}{t}\right)-22=0\)
Đặt \(t-\frac{2}{t}=a\Rightarrow t^2+\frac{4}{t^2}=a^2+4\)
\(\Rightarrow5\left(a^2+4\right)-3a-22=0\Leftrightarrow5a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{2}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t-\frac{2}{t}=1\\t-\frac{2}{t}=-\frac{2}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t^2-t-2=0\\t^2+\frac{2}{5}t-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\\t=\frac{\sqrt{51}-1}{5}\\t=\frac{-\sqrt{51}-1}{5}\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=2\\\sqrt{x^2+2}=\frac{\sqrt{51}-1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=\frac{2-2\sqrt{51}}{25}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=\pm\sqrt{2}\)