Multiply the numerator and the denominator by the conjugate of the denominator.

The given expression is $\frac{2+\sqrt{3}}{2-\sqrt{3}}$.
The conjugate of the denominator $2-\sqrt{3}$ is $2+\sqrt{3}$.
Multiply the fraction by $\frac{2+\sqrt{3}}{2+\sqrt{3}}$:

Numerator: $(2+\sqrt{3})(2+\sqrt{3}) = (2)^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$
Denominator: $(2-\sqrt{3})(2+\sqrt{3})$. This is in the form $(x-y)(x+y) = x^2 - y^2$.
So, $(2)^2 - (\sqrt{3})^2 = 4 - 3 = 1$.

The expression is in the form $a + b\sqrt{c}$, where $a=7$, $b=4$, and $c=3$.

The rationalized expression is $7 + 4\sqrt{3}$.