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

As I'm trying to study calculus so I will be thankfull to just a hint, not full solution.

Thanks.

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1 Answer 1

up vote 2 down vote accepted

Hint: Let $\sqrt{x}=y$. Chances are good you will find things so much more familiar that everything will become clear.

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Thanks. $\sqrt{x}=y$ and multiplying the first fraction on $\frac{a-1}{a-1}$ helped me. –  demas Feb 21 '13 at 6:18
    
I think of it this way. After cancelling the obvious $y$ we get $\frac{(y+1)(y^3-1)}{y^2+y+1}$. But $y^3-1$ factors as $(y-1)(y^2+y+1)$. Another cancellation. –  André Nicolas Feb 21 '13 at 6:34
    
Here is my solution: $\frac{a+1}{a(a^2+a+1)} \frac{a(a^3-1)}{1} = \frac{(a+1)(a-1)}{a(a^2+a+1)(a-1)} \frac{a(a^3-1)}{1}=\frac{(a^2-1)}{a(a^3-1)} \frac{a(a^3-1)}{1} = a^2 - 1$ Thanks :) –  demas Feb 21 '13 at 6:44
    
You multiplied instead of factoring. Multiplying is more fun. –  André Nicolas Feb 21 '13 at 6:47
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