If $x$ is positive, then why does $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$? Given that $x$ is positive, $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$
I've been trying to convert the left side of the equation to the right side: 
$$ \frac{1}{\sqrt{x+1} + \sqrt{x}}$$
But then how can I flip this round to be what I have on the right side? 
I know that $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, so I would think that this would give me
$$(x+1)^{-1/2} + x^{-1/2}$$
Which I thought would then convert to 
$$-\sqrt{x+1} + (-\sqrt{x}) $$
so I'm not sure how the first part $\sqrt{x+1}$ got to be positive

Clear the denominator to get:
$$1 = (\sqrt{x+1} + \sqrt{x})(\sqrt{x+1}-\sqrt{x}) = (x+1)-x.$$
As Tilper points out, this is not a correct proof per se, but once you see this, you can reverse the ordering of your logic and write a proper proof.
In this case it would be more proper to write:
$$1 = (x+1)-x = (\sqrt{x+1} + \sqrt{x})(\sqrt{x+1}-\sqrt{x}). $$
Therefore since $\sqrt{x+1} + \sqrt{x} \neq 0$,
$$\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}.$$
