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Someone please help! :) (x^2-4)/(x-8)/(x-2) (x^2+2x+1)/(x-2) over (x^2-1)/(x^2-4)
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factor your terms first
So (x+2)(x-2)(x-2)/x-8; etc
Wait what?? haha
1st part isit \[\frac{(x^2-4)}{\frac{x-8}{(x-2}}\]?
yes
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or \[\frac{(\frac{x^2-4}{x-8})}{x-2}\] it has their difference
the first one
so @treeman5823 has gave u the answer..... Let say 0.5=1/2 \[\frac{2}{\frac{1}{2}}\] 2/0.5=4 right? That how we move we move the 2 in the 1/2 to the numerator that's it
ohh ok
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