# Thread: simplify a rational fraction

1. ## simplify a rational fraction

I need to simplify the following fraction

$$\frac{2}{x^2-4}+\frac{1}{2x-x^2}$$

The answer should be $$\frac{1}{x(x+2)}$$

How do i preceed??

2. Originally Posted by strunz
I need to simplify the following fraction

$$\frac{2}{x^2-4}+\frac{1}{2x-x^2}$$

The answer should be $$\frac{1}{x(x+2)}$$

How do i preceed??
Use the fact that $x^2- 4= (x- 2)(x+ 2)$ and that $2x- x^2= x(2- x)= -x(x-2)$

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

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).

3. $\frac{2}{x^2-4}+\frac{1}{2x-x^2}=$

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

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

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

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

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

correct plz

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

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

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

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

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

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

correct plz
Hi!

Your mistake is in the last fraction:

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

Can you see why?

5. Yes. I see now. Thx