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mollyann

  • 3 years ago

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  1. mollyann
    • 3 years ago
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  2. Nurali
    • 3 years ago
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    i think A.

  3. jim_thompson5910
    • 3 years ago
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    \[\Large \frac{\frac{x^2+2x+1}{x-2}}{\frac{x^2-1}{x^2-4}}\] is the same as \[\Large \frac{x^2+2x+1}{x-2} \times \frac{x^2-4}{x^2-1}\]

  4. jim_thompson5910
    • 3 years ago
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    and you can combine the fractions to get \[\Large \frac{(x^2+2x+1)(x^2-4)}{(x-2)(x^2-1)}\]

  5. jim_thompson5910
    • 3 years ago
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    then factor everything to get \[\Large \frac{(x+1)(x+1)(x-2)(x+2)}{(x-2)(x-1)(x+1)}\]

  6. mollyann
    • 3 years ago
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    so what would the answer be ?

  7. Nurali
    • 3 years ago
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    A is correct lol.

  8. jim_thompson5910
    • 3 years ago
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    yes, cancelling the common factors leads to \[\Large \frac{(x+1)(x+2)}{x-1}\] which is A

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