Đặt: \(t=x\sqrt{2x^2+4}\Rightarrow t^2=2\left(x^4+2x^2\right)\Rightarrow x^2\left(x^2+2\right)=\frac{t^2}{2}\)
Ta được phương trình sau: \(\frac{t^2}{2}=4-t\Leftrightarrow t^2+2t-8=0\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=2\end{matrix}\right.\)
Với: \(t=-4\) ta được: \(x\sqrt{2x^2+4}=-4\Leftrightarrow\left\{{}\begin{matrix}x< 0\\2\left(x^4+2x^2\right)=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x^4+2x^2-8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x^2=2\end{matrix}\right.\Leftrightarrow x=-\sqrt{2}\)
Với: \(t=2\) ta được: \(x\sqrt{2x^2+4}=2\Leftrightarrow\left\{{}\begin{matrix}x>0\\2\left(x^4+2x^2\right)=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x^4+2x^2-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x^2=\sqrt{3}-1\end{matrix}\right.\Leftrightarrow x=\sqrt{\sqrt{x}-1}\)
Vậy ...........
