# simplify a rational fraction

• Sep 3rd 2009, 03:31 AM
strunz
simplify a rational fraction
I need to simplify the following fraction

[Math]\frac{2}{x^2-4}+\frac{1}{2x-x^2}[/tex]

How do i preceed??
• Sep 3rd 2009, 04:01 AM
HallsofIvy
Quote:

Originally Posted by strunz
I need to simplify the following fraction

[Math]\frac{2}{x^2-4}+\frac{1}{2x-x^2}[/tex]

How do i preceed??

Use the fact that $\displaystyle x^2- 4= (x- 2)(x+ 2)$ and that $\displaystyle 2x- x^2= x(2- x)= -x(x-2)$

That makes your fraction $\displaystyle \frac{2}{(x- 2)(x+ 2)}- \frac{1}{x(x-2)}$

The "common denominator" is (x- 2)(x+ 2)(x). Multiply numerator and denominator of the first fraction by x and the of the second fraction by (x+ 2).
• Sep 3rd 2009, 06:59 AM
strunz
$\displaystyle \frac{2}{x^2-4}+\frac{1}{2x-x^2}=$

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

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

$\displaystyle \frac{2x}{(x+2)(x-2)(x)}-\frac{x+2}{(x)(x-2)(x+2)}=$

$\displaystyle \frac{2x-(x+2)}{(x)(x-2)(x+2)}=$

$\displaystyle \frac{2x}{(x)(x-2)}$

correct plz
• Sep 3rd 2009, 08:08 AM
Defunkt
Quote:

Originally Posted by strunz
$\displaystyle \frac{2}{x^2-4}+\frac{1}{2x-x^2}=$

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

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

$\displaystyle \frac{2x}{(x+2)(x-2)(x)}-\frac{x+2}{(x)(x-2)(x+2)}=$

$\displaystyle \frac{2x-(x+2)}{(x)(x-2)(x+2)}=$

$\displaystyle \frac{2x}{(x)(x-2)}$

correct plz

Hi!

Your mistake is in the last fraction:

$\displaystyle \frac{2x-(x+2)}{x(x+2)(x-2)} \neq \frac{2x}{x(x-2)}$

Can you see why?
• Sep 3rd 2009, 08:31 AM
strunz
Yes. I see now. Thx