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I know this is a very basic question, but could someone please mathematically explain, why this is true:

$\sqrt{x} \cdot \frac{1}{x} = \frac{1}{\sqrt{x}}$

Wolfram|Alpha can confirm this.

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Let us suppose that $x$ is positive. If it is not, our expression is not defined. Note that essentially by definition, $x=\sqrt{x}\sqrt{x}.$
It follows that $\frac{\sqrt{x}}{x}=\frac{\sqrt{x}}{\sqrt{x}\sqrt{x}}=\frac{1}{\sqrt{x}}.$ (For the last step, we divided top and bottom by $\sqrt{x}$.)

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    @Robert Israel: True, the x>0 stuff could be omitted. But I wanted to remind the OP that, at the level (s)he is working, there are restrictions.2012-02-02
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$\sqrt{x}\cdot\frac{1}{x}= \color{Red}{\sqrt{x}}\cdot\frac{1}{\color{Red}{\sqrt{x}}\cdot\sqrt{x}}=\frac{1}{\sqrt{x}}.$

A more general method: $x^{1/2}\cdot x^{-1}=x^{1/2-1}=x^{-1/2}=(x^{1/2})^{-1}.$