simplify

• Jan 26th 2009, 11:58 AM
william
simplify
simplify

I) $\displaystyle \frac{x}{x-1}+\frac{x+2}{x^2-1}$
• Jan 26th 2009, 12:18 PM
Mush
Quote:

Originally Posted by william
simplify

I) $\displaystyle \frac{x}{x-1}+\frac{x+2}{x^2-1}$

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

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

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

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

$\displaystyle \frac{x^2+2x+ 2}{(x+1)(x-1)}$
• Jan 26th 2009, 12:24 PM
william
Quote:

Originally Posted by Mush
$\displaystyle \frac{x}{x-1}+\frac{x+2}{(x-1)(x+1)}$

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

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

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

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

very nicely done thanks(Nod)
• Jan 26th 2009, 02:43 PM
william
Quote:

Originally Posted by Mush
$\displaystyle \frac{x}{x-1}+\frac{x+2}{(x-1)(x+1)}$

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

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

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

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

what would change if it was I) $\displaystyle \frac{x}{x-1}+\frac{x+1}{x^2-1}$