K kpx001 Junior Member Joined Mar 6, 2006 Messages 119 Aug 2, 2007 #1 4x^2 / (x^2 - 4x + 4) / (2x) / (x^2 - 4) 4x^2 / (x-2)(x-2) / 2x /(x-2)(x+2) 4x^2 / (x-2)(x-2) times (x-2)(x+2)/(2x) 2x(x+2) / (x-2) = final?
4x^2 / (x^2 - 4x + 4) / (2x) / (x^2 - 4) 4x^2 / (x-2)(x-2) / 2x /(x-2)(x+2) 4x^2 / (x-2)(x-2) times (x-2)(x+2)/(2x) 2x(x+2) / (x-2) = final?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Aug 2, 2007 #2 You got it. Very good
D Deleted member 4993 Guest Aug 2, 2007 #3 Re: performing operation 4x^2 / (x^2 - 4x + 4) / (2x) / (x^2 kpx001 said: should be written as: [4x^2 / (x^2 - 4x + 4) ] / [(2x) / (x^2 - 4)] [4x^2 / {(x-2)(x-2)}] / [2x /{(x-2)(x+2)}] 4x^2 / [(x-2)(x-2)] * (x-2)(x+2)/(2x) 2x(x+2) / (x-2) = final? Click to expand...
Re: performing operation 4x^2 / (x^2 - 4x + 4) / (2x) / (x^2 kpx001 said: should be written as: [4x^2 / (x^2 - 4x + 4) ] / [(2x) / (x^2 - 4)] [4x^2 / {(x-2)(x-2)}] / [2x /{(x-2)(x+2)}] 4x^2 / [(x-2)(x-2)] * (x-2)(x+2)/(2x) 2x(x+2) / (x-2) = final? Click to expand...
