Question

Giải phương trình \(x^2\left(x^2+2\right)=4-x\sqrt{2x^2+4}\)

Answer 1

Đặt \(x\sqrt{2x^2+4}=a\)

\(\Rightarrow\frac{1}{2}a^2=4-a\Leftrightarrow a^2+2a-8=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=2\left(x>0\right)\\x\sqrt{2x^2+4}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(2x^2+4\right)=4\\x^2\left(2x^2+4\right)=16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-2=0\\x^4+2x^2-8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1\\x^2=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)