help please

Thanks Radix,      

I'll just take a look   

$$\\\frac{x^2-1}{x^2-3x+2}>0\\\\
\frac{(x-1)(x+1)}{(x-1)(x-2)}>0\\\\
$NOTE: x cannot be 1 or 2$\\\\
\frac{(x+1)}{(x-2)}>0\\\\
\frac{(x-2)(x+1)}{1}>0\\\
(x-2)(x+1)>0\\\\
consider\;\;y=(x-2)(x+1)\\\\
$This is a concave up parabola with roots at x=2 and x=-1$\\\\
$It will be above the x axis (positive) at the two ends$\\\\
$roughly sketch it to convince yourself$\\\\
$therefore true for$\\\\
x<-1\;\;\;and \;\;\;x>2$$

x^2-1 > x^2-3x+2      | -x^2

     -1 > -3x +2

     -3 > -3x                  |   * (-1)

  3 < 3x                        | : 3        =>   1 < x      =>       x > 1

I hope it is right !

Gruß radix !

thanks but the answer which is written n my book is:x<-1 or x>2