Đáp án:
$\begin{array}{l}
Dkxd:R\\
x + \sqrt {{x^2} + 12} = \dfrac{{24}}{{\sqrt {{x^2} + 12} }}\\
\Rightarrow x.\sqrt {{x^2} + 12} + {x^2} + 12 = 24\\
\Rightarrow 2.x.\sqrt {{x^2} + 12} + 2{x^2} + 24 = 48\\
\Rightarrow {x^2} + 12 + 2.x.\sqrt {{x^2} + 12} + {x^2} - 36 = 0\\
\Rightarrow \left( {{x^2} + 12} \right) + 2.x.\sqrt {{x^2} + 12} + {x^2} = 36\\
\Rightarrow {\left( {\sqrt {{x^2} + 12} + x} \right)^2} = 36\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {{x^2} + 12} + x = 6\\
\sqrt {{x^2} + 12} + x = - 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {{x^2} + 12} = 6 - x\left( {dk:x \le 6} \right)\\
\sqrt {{x^2} + 12} = - 6 - x\left( {dk:x \le - 6} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 12 = {x^2} - 12x + 36\left( {x \le 6} \right)\\
{x^2} + 12 = {x^2} + 12x + 36\left( {x \le - 6} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 2
\end{array}$
