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

Rationalize the denominator:
\begin{align}
  \frac{1}{\sqrt{x+1} + \sqrt{x}} \cdot \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1} - \sqrt{x}} &= \frac{\sqrt{x+1}-\sqrt{x}}{\underbrace{\left(\sqrt{x+1} + \sqrt{x}\right) \cdot \left(\sqrt{x+1} - \sqrt{x}\right)}_{\color{red}{\text{FOIL}}}} \\[0.3cm]
  &= \cdots
\end{align}
Some notes:
If you want to show that two things are equal to each other, it's better to manipulate only one of them than it is to start by saying they're equal to each other and then manipulating both sides to verify, etc.
Your idea of converting to $(x+1)^{-1/2} + x^{-1/2}$ won't work because, in general, $\displaystyle \frac{1}{a+b} \ne \frac{1}{a} + \frac{1}{b}$.
