Step 1: Rationalize the denominator of the given expression. To rationalize $\frac{\sqrt{3}}{2-\sqrt{5}}$, multiply the numerator and denominator by the conjugate of the denominator, which is $2+\sqrt{5}$. $$ \frac{\sqrt{3}}{2-\sqrt{5}} = \frac{\sqrt{3}}{2-\sqrt{5}} \times \frac{2+\sqrt{5}}{2+\sqrt{5}} $$ Step 2: Perform the multiplication. Multiply the numerators: $$ \sqrt{3}(2+\sqrt{5}) = 2\sqrt{3} + \sqrt{3}\sqrt{5} = 2\sqrt{3} + \sqrt{15} $$ Multiply the denominators using the difference of squares formula $(x-y)(x+y) = x^2 - y^2$: $$ (2-\sqrt{5})(2+\sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1 $$ Step 3: Simplify the expression. $$ \frac{2\sqrt{3} + \sqrt{15}}{-1} = -(2\sqrt{3} + \sqrt{15}) = -2\sqrt{3} - \sqrt{15} $$ Step 4: Compare the simplified expression with $a\sqrt{b} + c\sqrt{d}$. We have $-2\sqrt{3} - \sqrt{15}$. This can be written as $-2\sqrt{3} + (-1)\sqrt{15}$. Comparing this to $a\sqrt{b} + c\sqrt{d}$: $a = -2$ $b = 3$ $c = -1$ $d = 15$ All values $a, b, c, d$ are rational numbers. The values are: $a = \boxed{-2}$ $b = \boxed{3}$ $c = \boxed{-1}$ $d = \boxed{15}$ Send me the next one 📸