\(x^2\left(x^2+2\right)=4-x\sqrt{2x^2+4}\)
Đặt \(t=x\sqrt{2x^2+4}\)
Pttt: \(\dfrac{t^2}{2}=4-t\)
\(\Leftrightarrow t^2+2t-8=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)
TH1: \(t=2\Rightarrow x\sqrt{2x^2+4}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2\left(2x^2+4\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^4+2x^2-2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x^2=-1+\sqrt{3}\end{matrix}\right.\)(do \(x^2\ge0\)) \(\Rightarrow x=\sqrt{-1+\sqrt{3}}\)
TH2: \(t=-4\Rightarrow x\sqrt{2x^2+4}=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2\left(2x^2+4\right)=16\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^4+2x^2-8=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x^2=2\end{matrix}\right.\)(do \(x^2\ge0\))\(\Rightarrow x=-\sqrt{2}\)
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