<s>[INST] 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
 [/INST] One can basically use
$$
\frac{1}{a+b}=\frac{1}{a+b}\frac{a-b}{a-b}=\frac{a-b}{a^2-b^2}
$$
valid for $a-b\ne0$. If $a=\sqrt{x+1}$ and $b=\sqrt{x}$, you have
$$
a^2-b^2=(x+1)-x=1
$$
so
$$
\frac{1}{\sqrt{x+1}+\sqrt{x}}=
\frac{1}{a+b}=\frac{a-b}{a^2-b^2}=
\frac{\sqrt{x+1}-\sqrt{x}}{1}=
\sqrt{x+1}-\sqrt{x}
$$
With some practice, you'll do the steps automatically:
$$
\frac{1}{\sqrt{x+1}+\sqrt{x}}=
\frac{1}{\sqrt{x+1}+\sqrt{x}}\,
\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}=
\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}\,)^2-(\sqrt{x})^2}
$$
</s>