"rationalize" the denominator rather than the numerator

[powp]

Hello All

I get to thrown off when there is a question with radicals . Can you please help

Simplify the following

$$\frac{\sqrt{x} + 1}{\sqrt{x} - 1}$$


and Preform the indicated operations and simplfy


$$\sqrt{x}(\sqrt{x} + 1)(2\sqrt{x}-1)$$

Thanks

P

 

[Anzas]

the first:

$$\frac{\sqrt{x} + 1}{\sqrt{x} - 1}$$

$$\frac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} - 1)}$$

$$\frac{x-1}{x -2 \sqrt{x} +1}$$

the second:

$$\sqrt{x}(\sqrt{x} + 1)(2\sqrt{x}-1)$$

$$\sqrt{x}(2x+ \sqrt{x}-1)$$

$$2x \sqrt{x}+x- \sqrt{x}$$

 

[HallsofIvy]

I would have been inclined to "rationalize" the denominator rather than the numerator:
$$\frac{\sqrt{x}+1}{\sqrt{x}-1}= \frac{(\sqrt{x}+1)(\sqrt{x}+1)}{(\sqrt{x}-1})(\sqrt{x}+1)}= \frac{x+ 2\sqrt{x}+1}{x-1}$$

 

[powp]

thanks that's the answers I got but was not sure

When would you "rationalize" the denominator over the numerator or the other way around.

 

[snoble]

Generally simplified form means always "rationalizing" the denominator. Also you can simplify a little bit more. $$\frac{x+ 2\sqrt{x}+1}{x-1} = \frac{ (1+\sqrt{x})^2}{x-1}$$