Sửa đề:
\(x^2+5x+1=\left(x+5\right)\sqrt{x^2+1}\)
\(\Leftrightarrow x^2+1+5x=\left(x+5\right)\sqrt{x^2+1}\)
\(\Leftrightarrow x^2+1+5x-x\sqrt{x^2+1}-5\sqrt{x^2+1}=0\)
\(\Leftrightarrow\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)+5\left(x-\sqrt{x^2+1}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\\\sqrt{x^2+1}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\left(không-có-x-tm\right)\\x^2+1=25\end{matrix}\right.\Leftrightarrow x=\pm\sqrt{24}\)
Vậy nghiệm của pt là \(x=\pm\sqrt{24}\)
