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Which of the following is equal to \(\frac{(11!)^2 - (9!)^2}{(11! - 9!)^2}\)

A. 1/10

B. 109/111

C. 1

D. 111/109

E. 10

Attachment:
2024-01-25_13-21-50.png

    \(\frac{(11!)^2 - (9!)^2}{(11! - 9!)^2}=\)\(=\frac{(11! - 9!)(11! + 9!)}{(11! - 9!)^2}=\)\(=\frac{11! + 9!}{11! - 9!}=\)\(=\frac{9!(10*11 + 1)}{9!(10*11 - 1)}=\)\(=\frac{10*11 + 1}{10*11 - 1}=\)\(=\frac{111}{109}.\)

Answer: D.
­
Can you explain why you did this step? 9!(10*11+1)/9!(10*11-1)
The step 9!(10*11+1)/9!(10*11-1) helps simplify the equation. You are taking the GCF from the numerator and denominator which both happen to be 9!. This allows you to cross out the 9! and only worry about the (10*11+1)/(10*11-1) portion.
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