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

You need to use a little trick with fractions. Recall that for any number $n$ that is not zero, $\frac{n}n=1$, and 1 times any number is itself.
So start with the left side: $\sqrt{1+x}-\sqrt{x}$. If we choose our $n$ to be $\sqrt{1+x}+\sqrt{x}$, then:
$$\sqrt{1+x}-\sqrt{x}=\frac{\sqrt{1+x}-\sqrt{x}}1\cdot\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}$$
Then "FOIL" the top:
$$\frac{\sqrt{1+x}^2+\sqrt{x}\sqrt{1+x}-\sqrt{x}\sqrt{1+x}-\sqrt{x}^2}{\sqrt{1+x}+\sqrt{x}}=\frac{1+x-x}{\sqrt{1+x}+\sqrt{X}}=\frac1{\sqrt{1+x}+\sqrt{x}}$$
