anonymous
  • anonymous
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Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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Nurali
  • Nurali
i think A.
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{\frac{x^2+2x+1}{x-2}}{\frac{x^2-1}{x^2-4}}\] is the same as \[\Large \frac{x^2+2x+1}{x-2} \times \frac{x^2-4}{x^2-1}\]

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jim_thompson5910
  • jim_thompson5910
and you can combine the fractions to get \[\Large \frac{(x^2+2x+1)(x^2-4)}{(x-2)(x^2-1)}\]
jim_thompson5910
  • jim_thompson5910
then factor everything to get \[\Large \frac{(x+1)(x+1)(x-2)(x+2)}{(x-2)(x-1)(x+1)}\]
anonymous
  • anonymous
so what would the answer be ?
Nurali
  • Nurali
A is correct lol.
jim_thompson5910
  • jim_thompson5910
yes, cancelling the common factors leads to \[\Large \frac{(x+1)(x+2)}{x-1}\] which is A

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