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