If 1/(3-"√"5) - (2+"√"5)/(3+"√"5) = a + b"√"c, find the value of a, b and c.

To rationalize the denominator, multiply the numerator and denominator by the conjugate of $3-\sqrt{5}$, which is $3+\sqrt{5}$.

$$\frac{1}{3-\sqrt{5}} = \frac{1}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}}$$ $$= \frac{3+\sqrt{5}}{(3)^2 - (\sqrt{5})^2}$$ $$= \frac{3+\sqrt{5}}{9 - 5}$$ $$= \frac{3+\sqrt{5}}{4}$$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of $3+\sqrt{5}$, which is $3-\sqrt{5}$.

$$\frac{2+2\sqrt{5}}{3+\sqrt{5}} = \frac{2+2\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}$$ $$= \frac{(2+2\sqrt{5})(3-\sqrt{5})}{(3)^2 - (\sqrt{5})^2}$$

Expand the numerator:

$$(2+2\sqrt{5})(3-\sqrt{5}) = 2(3) - 2(\sqrt{5}) + 2\sqrt{5}(3) - 2\sqrt{5}(\sqrt{5})$$ $$= 6 - 2\sqrt{5} + 6\sqrt{5} - 2(5)$$ $$= 6 + 4\sqrt{5} - 10$$ $$= -4 + 4\sqrt{5}$$

The denominator is $9-5=4$.
So, the second term becomes:

$$\frac{-4+4\sqrt{5}}{4} = \frac{4(-1+\sqrt{5})}{4}$$ $$= -1+\sqrt{5}$$

$$\frac{3+\sqrt{5}}{4} - (-1+\sqrt{5}) = a+b\sqrt{c}$$ $$\frac{3+\sqrt{5}}{4} + 1 - \sqrt{5} = a+b\sqrt{c}$$

Find a common denominator, which is 4.

$$\frac{3+\sqrt{5}}{4} + \frac{4}{4} - \frac{4\sqrt{5}}{4} = a+b\sqrt{c}$$ $$\frac{3+\sqrt{5} + 4 - 4\sqrt{5}}{4} = a+b\sqrt{c}$$

Combine the rational parts and the irrational parts:

$$\frac{(3+4) + (\sqrt{5} - 4\sqrt{5})}{4} = a+b\sqrt{c}$$ $$\frac{7 - 3\sqrt{5}}{4} = a+b\sqrt{c}$$

$$\frac{7}{4} - \frac{3\sqrt{5}}{4} = a+b\sqrt{c}$$

This can be written as:

$$\frac{7}{4} + \left(-\frac{3}{4}\right)\sqrt{5} = a+b\sqrt{c}$$

By comparing the rational parts:
$a = \frac{7}{4}$
By comparing the coefficients of $\sqrt{5}$:
$b = -\frac{3}{4}$
By comparing the radicands:
$c = 5$

The values are:
$a = \boxed{\frac{7}{4}}$
$b = \boxed{-\frac{3}{4}}$
$c = \boxed{5}$