\(x^4+\left(x^2+1\right)\sqrt{x^2+1}-1=0\)
Đặt \(\left\{{}\begin{matrix}x^2+1=a\\x^2-1=b\end{matrix}\right.\) \(\Leftrightarrow a-b=2\Leftrightarrow b=a-2\)
pt \(\Leftrightarrow ab+a\sqrt{a}=0\)
\(\Leftrightarrow a\left(a-2\right)+a\sqrt{a}=0\)
\(\Leftrightarrow a^2+a\sqrt{a}-2a=0\)
\(\Leftrightarrow a\left(a+\sqrt{a}-2\right)=0\)
\(\Leftrightarrow a\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x^2+1=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\varnothing\\x=0\end{matrix}\right.\)
Vậy \(x=0\) là nghiệm duy nhất của pt.
