Answer
$\dfrac{-7-\sqrt{21}}{4}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the denominator of the given expression, $
\dfrac{\sqrt{7}}{\sqrt{3}-\sqrt{7}}
,$ 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{3}+\sqrt{7}
.$ Multiplying the conjugate to both the numerator and the denominator of the expression above results to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{7}}{\sqrt{3}-\sqrt{7}}\cdot\dfrac{\sqrt{3}+\sqrt{7}}{\sqrt{3}+\sqrt{7}}
.\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{7(3)}+\sqrt{7(7)}}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})}
\\\\=
\dfrac{\sqrt{21}+\sqrt{7^2}}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})}
\\\\=
\dfrac{\sqrt{21}+7}{(\sqrt{3}-\sqrt{7})(\sqrt{3}+\sqrt{7})}
.\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{21}+7}{(\sqrt{3})^2-(\sqrt{7})^2}
\\\\=
\dfrac{\sqrt{21}+7}{3-7}
\\\\=
\dfrac{\sqrt{21}+7}{-4}
\\\\=
-\dfrac{\sqrt{21}+7}{4}
\\\\=
\dfrac{-\sqrt{21}-7}{4}
\\\\=
\dfrac{-7-\sqrt{21}}{4}
.\end{array}
