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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?

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There is nothing wrong with your computation. Actually, it's the rearrangement of the terms in the denominator you are confused about. Remember that addition of two terms is always commutative, for instance $1 + \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{x}} + 1$. Actually, it's the second radicand that messes up the computation, therefore making the answer incorrect. – NasuSama May 12 '13 at 13:51

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.

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$\sqrt{x \pm \sqrt x} = \sqrt{x (1\pm\frac {\sqrt x}{x}}) = \sqrt x \sqrt{1\pm\frac {1}{\sqrt x}}$

Looks like you swapped the terms on both sides of the negative $-$ sign. It is okay for the $+$ but not for the $-$.

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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.

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$$ \frac{(x + \sqrt{x}) - (x - \sqrt{x} )}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}} = \frac{2 \sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}}$$ 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.

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