Nhận thấy \(x=0\) không phải nghiẹm của pt, chia 2 vế cho \(x^4\):
\(\Leftrightarrow5-\frac{2}{x^2}-3\sqrt{\frac{1}{x^2}+\frac{2}{x^4}}=\frac{4}{x^4}\)
\(\Leftrightarrow2\left(\frac{1}{x^2}+\frac{2}{x^4}\right)+3\sqrt{\frac{1}{x^2}+\frac{2}{x^4}}-5=0\)
Đặt \(\sqrt{\frac{1}{x^2}+\frac{2}{x^4}}=a\ge0\)
\(\Rightarrow2a^2+3a-5=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{5}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{1}{x^2}+\frac{2}{x^4}}=1\Leftrightarrow\frac{2}{x^4}+\frac{1}{x^2}-1=0\)
Đặt \(\frac{1}{x^2}=t\) (\(t>0\)) \(\Rightarrow2t^2+t-1=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{x^2}=\frac{1}{2}\Rightarrow x=\pm\sqrt{2}\)
