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calyne

show me how to solve for x when 1/(x+1)^2 = [x/(x+1) - 2]/(x-1)

  • 2 years ago
  • 2 years ago

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  1. satellite73
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    bunch of algebra, i will try to write it

    • 2 years ago
  2. calyne
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    ik thanks

    • 2 years ago
  3. satellite73
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    \[\frac{x}{x+1}-2=\frac{-x-2}{x+1}\] is a start

    • 2 years ago
  4. satellite73
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    so right hand sides is \[\frac{-x-2}{(x+1)(x-1)}\]

    • 2 years ago
  5. calyne
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    wait how did you start out like that

    • 2 years ago
  6. satellite73
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    \[\frac{1}{(x+1)^2}=\frac{-x-2}{(x+1)(x-1)}\] \[\frac{1}{x+1}=\frac{-x-2}{x-1}\] \[x-1=(x+1)(-x-2)\] \[x-1=-x^2-3x-2\] \[x^2+3x+1=0\]

    • 2 years ago
  7. satellite73
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    \[\frac{x}{x+1}-2=\frac{x}{x+1}-\frac{2(x+1)}{x+1}\] etc

    • 2 years ago
  8. calyne
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    still don't get thte first step talk about moronic what the flutter

    • 2 years ago
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