How do you simplify #(d+7)/(d^-49)#?

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Answer 1 · 2015-07-03T19:05:25

I think you meant #(d+7)/(d^2-49)#

If so, the answer is #1/(d-7)# with exclusion #d != -7#

#(d+7)/(d^2-49)#

#=(d+7)/(d^2-7^2)#

#=(d+7)/((d-7)(d+7))#

#=1/(d-7)#

With exclusion #d != -7#

using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=d# and #b=7#