Answer
$\frac{\sqrt{3}-\sqrt {7}}{\sqrt {3}-\sqrt {7}}$.
Work Step by Step
The conjugate of $\sqrt {3}+\sqrt {7}$ is $\sqrt {3}-\sqrt {7}$.
Appropriate form of $1$ that can be used as a factor to eliminate the radical in the denominator is $\frac{\sqrt{3}-\sqrt {7}}{\sqrt {3}-\sqrt {7}}$.
The denominator becomes:
$(\sqrt 3+\sqrt 7)(\sqrt 3-\sqrt 7)=(\sqrt 3)^2-(\sqrt 7)^2=3-7=-4$