#### Answer

$\dfrac{\sqrt{x}+\sqrt{5}}{x-5}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{1}{\sqrt{x}-\sqrt{5}}
,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result.
$\bf{\text{Solution Details:}}$
Multiplying the numerator and the denominator by the conjugate of the denominator results to
\begin{array}{l}\require{cancel}
\dfrac{1}{\sqrt{x}-\sqrt{5}}\cdot\dfrac{\sqrt{x}+\sqrt{5}}{\sqrt{x}+\sqrt{5}}
\\\\=
\dfrac{1(\sqrt{x}+\sqrt{5})}{(\sqrt{x})^2-(\sqrt{5})^2}
\\\\=
\dfrac{\sqrt{x}+\sqrt{5}}{(\sqrt{x})^2-(\sqrt{5})^2}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{x}+\sqrt{5}}{x-5}
.\end{array}