Đặt $t=\sqrt{x^2+5}(t\ge 0)$
$\begin{array}{l}
\Rightarrow {x^2} = {t^2} - 5\\
PT \Leftrightarrow {t^2} - 5 - 5t - 5 = 0\\
\Leftrightarrow {t^2} - 5t - 10 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = \dfrac{{5 + \sqrt {65} }}{2}(TM)\\
t = \dfrac{{5 - \sqrt {65} }}{2}(L)
\end{array} \right.\\
\Rightarrow \sqrt {{x^2} + 5} = \dfrac{{5 + \sqrt {65} }}{2} \Leftrightarrow {x^2} + 5 = \dfrac{{90 + 10\sqrt {65} }}{4}\\
\Leftrightarrow {x^2} = \dfrac{{35 + 5\sqrt {65} }}{2} \Leftrightarrow x = \pm \sqrt {\dfrac{{35 + 5\sqrt {65} }}{2}}
\end{array}$
