Answer
$\dfrac{13-3\sqrt[]{21}}{5}$
Work Step by Step
Multiplying both the numerator and the denominator by $
2\sqrt[]{3}+\sqrt{7}
$, then the rationalized-denominator form of the given expression, $
\dfrac{\sqrt[]{3}-\sqrt{7}}{2\sqrt[]{3}+\sqrt{7}}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt[]{3}-\sqrt{7}}{2\sqrt[]{3}+\sqrt{7}}\cdot\dfrac{2\sqrt[]{3}-\sqrt{7}}{2\sqrt[]{3}-\sqrt{7}}
\\\\=
\dfrac{\sqrt[]{3}(2\sqrt[]{3})+\sqrt[]{3}(-\sqrt{7})-\sqrt{7}(2\sqrt[]{3})-\sqrt{7}(-\sqrt{7})}{(2\sqrt[]{3})^2-(\sqrt{7})^2}
\\\\=
\dfrac{2\sqrt[]{3(3)}-\sqrt[]{3(7)}-2\sqrt[]{3(7)}+\sqrt{7(7)}}{4(3)-7}
\\\\=
\dfrac{2\sqrt[]{(3)^2}-\sqrt[]{21}-2\sqrt[]{21}+\sqrt{(7)^2}}{12-7}
\\\\=
\dfrac{2(3)-\sqrt[]{21}-2\sqrt[]{21}+7}{12-7}
\\\\=
\dfrac{6-\sqrt[]{21}-2\sqrt[]{21}+7}{5}
\\\\=
\dfrac{(6+7)+(-\sqrt[]{21}-2\sqrt[]{21})}{5}
\\\\=
\dfrac{13-3\sqrt[]{21}}{5}
.\end{array}