# Any guideline??

• Apr 22nd 2009, 05:29 PM
Alienis Back
Any guideline??
The problem
$(1 - x^-2) : (1 - x^-1)$

My aproach:

= $(\frac{1}{1} - \frac{1}{x^2}) : (\frac{1}{1} - \frac{1}{x})$

= $(\frac{x^2}{x^2} - \frac{1}{x^2}) : (\frac{x}{x} - \frac{1}{x})$

= $\frac{x^2 - 1}{x^2} : \frac{x - 1}{x}$

= $\frac{x^2 - 1}{x^2} * \frac{x}{x - 1}$

= $\frac{x^3 - x}{x^3 - x^2}$

But according to the book the answer should be = $\frac{x + 1}{x}$

Mmm...-2 and -1 on the original problem should be exponents.
• Apr 22nd 2009, 05:34 PM
Prove It
Quote:

Originally Posted by Alienis Back
The problem
$(1 - x^-2) : (1 - x^-1)$

My aproach:

= $(\frac{1}{1} - \frac{1}{x^2}) : (\frac{1}{1} - \frac{1}{x})$

= $(\frac{x^2}{x^2} - \frac{1}{x^2}) : (\frac{x}{x} - \frac{1}{x})$

= $\frac{x^2 - 1}{x^2} : \frac{x - 1}{x}$

= $\frac{x^2 - 1}{x^2} * \frac{x}{x - 1}$

= $\frac{x^3 - x}{x^3 - x^2}$

But according to the book the answer should be = $\frac{x + 1}{x}$

Mmm...-2 and -1 on the original problem should be exponents.

You haven't gone far enough.

$\frac{x^3 - x}{x^3 - x^2}$

$= \frac{x(x^2 - 1)}{x^2(x - 1)}$

$= \frac{x^2 - 1}{x(x - 1)}$

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

$= \frac{x + 1}{x}$.
• Apr 22nd 2009, 05:45 PM
Alienis Back
What did you do here?

Quote:

Originally Posted by Prove It
You haven't gone far enough.

$= \frac{x(x^2 - 1)}{x^2(x - 1)}$

$= \frac{x^2 - 1}{x(x - 1)}$.

I mean $x:x^2 = x^1-2 = x^-1$...mmm...
• Apr 22nd 2009, 07:36 PM
Prove It
Quote:

Originally Posted by Alienis Back
What did you do here?

I mean $x:x^2 = x^1-2 = x^-1$...mmm...

[tex]Remember that $x^2 = x\cdot x$

So $\frac{x}{x^2} = \frac{x}{x\cdot x} = \frac{1}{x}$.

Therefore

$\frac{x(x^2 - 1)}{x^2(x - 1)} = \frac{x(x^2 - 1)}{x\cdot x(x - 1)} = \frac{1(x^2 - 1)}{x(x - 1)}$