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$$ \cfrac{ \;\cfrac{\sqrt{x} + 1}{1+ \sqrt{x} + x} \;}{ \cfrac{1}{x^2 - \sqrt{x}} } $$

I'm not sure if I'm missing anything... other expressions are easy, but I'm not sure what to do with this one.

EDIT: Thanks for doing it for me, I'm not familiar with LaTeX program.

EDIT2: Okay, after searching through the book, there is a solution, but without steps:

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

So now, can someone explain? Maybe there's an error...

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No, no, no! Use LaTeX to write mathematics in this site, in particular something that may so easily be misunderstood... – DonAntonio Nov 10 '13 at 16:16
I fixed up the MathJax/LaTex notation as best I could, but the colon in the middle suggests a compound fraction or proportion? – hardmath Nov 10 '13 at 16:18
Suggests division. ( ':' is '/' ) Now you see why I left it as ':', because of the size... Now just leave it as it is, if a soul is willing to help, they'll manage. – Guest114 Nov 10 '13 at 16:19
Please review my changes for accuracy! – hardmath Nov 10 '13 at 16:20
Is there any typo? This cannot be further simplified! – freak_warrior Nov 10 '13 at 16:21
up vote 4 down vote accepted

Another way to simplify is to use this very basic rule: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ for the part $x+\sqrt{x}+1$. Indeed, $$\frac{\sqrt{x} + 1}{1+ \sqrt{x} + x}=\frac{(\sqrt{x} + 1)(\sqrt{x}-1)}{(1+ \sqrt{x} + x)(\sqrt{x}-1)}=\frac{x-1}{x\sqrt{x}-1}$$

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Thank you very much. The next step is to multiply it with conjugated form of denominator... and everything then goes well, and matches the original solution. – Guest114 Nov 10 '13 at 16:59
@amWhy: Thanks a lot. I wish someday I could get another Gold badge as you and Amzoti. I hope so. :-) – Babak S. Nov 10 '13 at 17:17
@B.S.: You will get much more gold my friend! +1 – Amzoti Nov 13 '13 at 14:22
Also, you and @amWhy already have a lot of gold in your real lives as you spend so much time helping others on the site. That is the best kid of gold there is as it is embedded in the heart! – Amzoti Nov 13 '13 at 14:26
@Amzoti: Amy left me nothing to express something. :-) – Babak S. Nov 13 '13 at 14:34

Hint: Use $$ \frac{\sqrt{x} + 1}{1 + \sqrt{x} + x} \div \frac{1}{x^2 - \sqrt{x}} = \frac{\sqrt{x} + 1}{1 + \sqrt{x} + x} \times \frac{x^2 - \sqrt{x}}{1} = \frac{- \sqrt{x} - x + x^2 + x^2\sqrt{x}}{{1 + \sqrt{x} + x}}. $$ Is the numerator a multiple of the denominator?

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I got to that step... but what's next? – Guest114 Nov 10 '13 at 16:25
You have to multiply the denominator by $x \sqrt{x}$ to get $x^2\sqrt{x}$ (in the numerator). What does multiplying by $x \sqrt{x}$ do to the rest of the denominator? – Math Student 020 Nov 10 '13 at 16:29
I'm more confused now, do I have to multiply the expression with xroot(x)/xroot(x) ? Why? – Guest114 Nov 10 '13 at 16:35
@Guest114, have you tried long division with the last form above? – dfeuer Nov 10 '13 at 17:01
This problem has been answered above. – Guest114 Nov 10 '13 at 17:04

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