Negative exponents

• Dec 1st 2008, 11:08 AM
DHS1
Negative exponents
Hey guys.

I know that x^-2 is equal to 1/x^2

But, this problem looks like this. It asks me to simplify this (AKA get rid of the negative exponents):

((x^-1)+(y^-1))/((x^-2)-(y^-2))

For some reason I don't think I can just flip the problem around to make it look like this

((x^2)-(y^2))/((x^1)+(y^1))

But if I can, can you let me know why? Thanks guys!
• Dec 1st 2008, 11:13 AM
Moo
Hello,

No you can't, because it would mean that :

$\frac{1}{x^{-2}-y^{-2}}=x^2-y^2$, which is similar to saying that $\frac{1}{a-b}=\frac 1a-\frac 1b$ which are both false.

So let's write your problem :

$\frac{x^{-1}+y^{-1}}{x^{-2}-y^{-2}}$

You can see that $x^{-2}=(x^{-1})^2$ and $y^{-2}=(y^{-1})^2$

Can you see a difference of two squares below ???

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Another way :

eliminate the negative coefficients.
In order to do that, multiply by $\frac{x^2y^2}{x^2y^2}$ :

$=\frac{xy^2+yx^2}{y^2-x^2}=\frac{xy(y+x)}{(y-x)(y+x)}=\frac{xy}{y-x}$
• Dec 1st 2008, 01:32 PM
DHS1
This is another problem that needs to be simplified.

It's going to look messy but I'll type it out right. Can someone tell me how to make it look like it does on the paper?

Anyways:

EDIT: sqrt = square root

(((2x)/(sqrt(x-1)))-sqrt(x-1))/(x-1)

Again sorry for the sloppiness >_<
• Dec 1st 2008, 01:42 PM
masters
Quote:

Originally Posted by DHS1
This is another problem that needs to be simplified.

It's going to look messy but I'll type it out right. Can someone tell me how to make it look like it does on the paper?

Anyways:

(((2x)/(sqrt(x-1)))-sqrt(x-1))/(x-1)

Again sorry for the sloppiness >_<

You'll have to tell me if this is what you have coded:

$\dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1}$
• Dec 1st 2008, 01:50 PM
DHS1
Quote:

Originally Posted by masters
You'll have to tell me if this is what you have coded:

$\dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1}$

Yeah it is. Did you actually type \dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1} between the math brackets? Or do you use a program and then copy/paste or something o_O;

Thanks.
• Dec 1st 2008, 02:12 PM
masters
Quote:

Originally Posted by DHS1
Yeah it is. Did you actually type \dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1} between the math brackets? Or do you use a program and then copy/paste or something o_O;

Thanks.

Just type it in just like that. You can go here for a tutorial on this tool.

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

$\dfrac{x+1}{(x-1)(\sqrt{x-1})} \cdot \dfrac{\sqrt{x-1}}{\sqrt{x-1}}=\dfrac{(x+1)(\sqrt{x-1})}{(x-1)^2}$
• Dec 1st 2008, 02:26 PM
DHS1
Wow great! I'm learning so much. Last one I'll ask for help with, I promise!

Directions: Re-write each expression as a single fraction in lowest terms.

$\frac{3}{a}+\frac{2}{a^2}-\frac{2}{a-1}$

EDIT: My problem is I don't know how to find out what the common denominator is. How can I go about doing that?

Thanks!
• Dec 2nd 2008, 12:18 AM
earboth
Quote:

Originally Posted by DHS1
Wow great! I'm learning so much. Last one I'll ask for help with, I promise!

Directions: Re-write each expression as a single fraction in lowest terms.

$\frac{3}{a}+\frac{2}{a^2}-\frac{2}{a-1}$

EDIT: My problem is I don't know how to find out what the common denominator is. How can I go about doing that?

Thanks!

how to find out what the common denominator is

1. Factorize each denominator.

2. The common denominator must contain the original denominators as a factor:

$d_1=a$
$d_2=a \cdot a$
$d_3=(a-1)$

Thus the common denominator is $D=a\cdot a\cdot (a-1)$

$\dfrac{3}{a}+\dfrac{2}{a^2}-\dfrac{2}{a-1} = \dfrac{3a(a-1)}{a^2(a-1)}+\dfrac{2(a-1)}{a^2(a-1)}-\dfrac{2a^2}{a^2(a-1)} =$ $\dfrac{3a^2-3a+2a-2-2a^2}{a^2(a-1)}= \dfrac{a^2-a-2}{a^2(a-1)} = \dfrac{(a-2)(a+1)}{a^2(a-1)}$

which can't be simplified any more.

By the way: Do yourself and do us a favour and start a new thread if you have a new question. Otherwise you'll risk that nobody will notice that you are in need of some help.
• Dec 2nd 2008, 08:09 AM
DHS1
Quote:

Originally Posted by earboth
how to find out what the common denominator is

1. Factorize each denominator.

2. The common denominator must contain the original denominators as a factor:

$d_1=a$
$d_2=a \cdot a$
$d_3=(a-1)$
Thus the common denominator is $D=a\cdot a\cdot (a-1)$
$\dfrac{3}{a}+\dfrac{2}{a^2}-\dfrac{2}{a-1} = \dfrac{3a(a-1)}{a^2(a-1)}+\dfrac{2(a-1)}{a^2(a-1)}-\dfrac{2a^2}{a^2(a-1)} =$ $\dfrac{3a^2-3a+2a-2-2a^2}{a^2(a-1)}= \dfrac{a^2-a-2}{a^2(a-1)} = \dfrac{(a-2)(a+1)}{a^2(a-1)}$