Question

\(\left(x^2+1\right)^2=5-x\sqrt{2x^2+4}\)

Answer 1

TXĐ: \(D=R\)

Đặt \(x\sqrt{x^2+2}=y\)

\(pt\Leftrightarrow x^4+2x^2+1=5-\sqrt{2}.x\sqrt{x^2+2}\)

\(\Leftrightarrow y^2+\sqrt{2}.y-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-2\sqrt{2}\\y=\sqrt{2}\end{matrix}\right.\)

TH1: \(y=-2\sqrt{2}\Leftrightarrow x\sqrt{x^2+2}=-2\sqrt{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(x^2+2\right)=8\\x< 0\end{matrix}\right.\Leftrightarrow x=-\sqrt{2}\)

TH2: \(y=\sqrt{2}\Leftrightarrow x\sqrt{x^2+2}=\sqrt{2}\)

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

Vậy phương trình có nghiệm \(x=\sqrt{-1+\sqrt{3}};x=-\sqrt{2}\)