ĐKXĐ: \(x\ge-1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\) ta được:
\(a^2-2b^2=ab\)
\(\Leftrightarrow a^2-ab-2b^2=0\Leftrightarrow\left(a+b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow a-2b=0\Leftrightarrow a=2b\)
\(\Leftrightarrow a^2=4b^2\Leftrightarrow x^2+1=4\left(x+1\right)\)
\(\Leftrightarrow x^2-4x-3=0\) (casio)