Answer
$\dfrac{\sqrt{15}-3}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the denominator of the given expression, $
\dfrac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}
,$ multiply the numerator and the denominator by the conjugate of the denominator.
$\bf{\text{Solution Details:}}$
Reversing the operator between the terms of the denominator in the expression above, the conjugate of the denominator is $
\sqrt{5}-\sqrt{3}
.$ Multiplying the conjugate to both the numerator and the denominator of the expression above results to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}\cdot\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{3(5)}-\sqrt{3(3)}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}
\\\\=
\dfrac{\sqrt{15}-\sqrt{3^2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}
\\\\=
\dfrac{\sqrt{15}-3}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}
.\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{15}-3}{(\sqrt{5})^2-(\sqrt{3})^2}
\\\\=
\dfrac{\sqrt{15}-3}{5-3}
\\\\=
\dfrac{\sqrt{15}-3}{2}
.\end{array}
