$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$? 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?
 A: You have
$$
\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}} = \sqrt{x(1 + \frac{1}{\sqrt{x}})} + \sqrt{x( 1 - \frac{1}{\sqrt{x}})}
$$
It looks like you only factored out a $\sqrt{x}$ from last term.
A: $\sqrt{x \pm \sqrt x} = \sqrt{x (1\pm\frac {\sqrt x}{x}}) = \sqrt x \sqrt{1\pm\frac {1}{\sqrt x}}$
Looks like you swapped the terms on both sides of the negative $-$ sign. It is okay for the $+$ but not for the $-$.
A: 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.
A: $$ \frac{(x + \sqrt{x}) - (x - \sqrt{x} )}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}} = \frac{2 \sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}}$$
Divide by $\sqrt{x}$ throughout. It simplifies to
$$\frac{2}{\dfrac{\sqrt{x + \sqrt{x}} }{\sqrt{x}} + \dfrac{\sqrt{x - \sqrt{x}}}{\sqrt{x}}}.$$
Now the denominator simplifies to
$\sqrt{\dfrac{x + \sqrt{x}}{x}} + \sqrt{\dfrac{x - \sqrt{x}}{x}} $ and the answer comes out nicely.
