$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$? According to an example in my text book:
$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$$
I don't see how this works. The closest I can get is:
$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2\sqrt{x}}{\sqrt{x}(\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1})} = \frac{2}{\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1}}$$
Which is slightly off. But I'm not even sure if I calculated that correctly. What am I missing, what's the way to think?

Use this equalities
$$(x + \sqrt{x}) - (x-\sqrt{x})=2\sqrt{x}$$
and
$$\sqrt{x+\sqrt{x}}=\sqrt{x(1+\frac{1}{\sqrt{x}})}=\sqrt{x}\sqrt{1+\frac{1}{\sqrt{x}}}$$
and the same result for
$$\sqrt{x-\sqrt{x}}$$
then simplify.
