ĐKXĐ: ...
\(\Leftrightarrow\left(x^4-2\right)\sqrt{x^2-1}-x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(x^4-2-x^2\sqrt{x^2-1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^4-2-x^2\sqrt{x^2-1}=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(\sqrt{x^2-1}=t\ge0\Rightarrow x^2=t^2+1\)
\(\left(t^2+1\right)^2-2-t\left(t^2+1\right)=0\)
\(\Leftrightarrow t^4-t^3+2t^2-t-1=0\)
\(\Leftrightarrow\left(t-1\right)\left(t^3+2t+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t^3+3t+1=0\end{matrix}\right.\) (pt dưới vô nghiệm do \(t\ge0\Rightarrow t^3+3t+1\ge1\))
