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chicagochica5 Group Title

Divide.

  • one year ago
  • one year ago

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  1. chicagochica5 Group Title
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    • one year ago
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  2. miteshchvm Group Title
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    the x^2 + 2x + 1 is thw square of (x + 1)^2 as per the previous questions you can solve it

    • one year ago
  3. chicagochica5 Group Title
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    The first part would be (x -2) 2x? + 1

    • one year ago
  4. theredhead1617 Group Title
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    Yes!

    • one year ago
  5. chicagochica5 Group Title
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    @theredhead1617 do you know how to solve the problem

    • one year ago
  6. miteshchvm Group Title
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    \[\frac{ \frac{ (x+1)(x+1) }{ x-2 } }{ \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\] \[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\]\] can you solv further?

    • one year ago
  7. chicagochica5 Group Title
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    No @miteshchvm I dont understand how to. I wish to know where to begin but i do not

    • one year ago
  8. miteshchvm Group Title
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    can you factorise x^2 + 2x + 1?

    • one year ago
  9. miteshchvm Group Title
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    x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) x^2 - 1 = (x +1)(x - 1) x^2 - 4 = (x-2)(x+2) you get it?

    • one year ago
  10. chicagochica5 Group Title
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    (x + 1) to second power?

    • one year ago
  11. miteshchvm Group Title
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    yes

    • one year ago
  12. miteshchvm Group Title
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    \[x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) \] \[x^2 - 1 = (x +1)(x - 1) \] \[x^2 - 4 = (x-2)(x+2)\] you get it? i wrote this instead of your question in order to simplify,

    • one year ago
  13. chicagochica5 Group Title
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    just a little bit, its hard

    • one year ago
  14. miteshchvm Group Title
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    now refer to my first comment

    • one year ago
  15. chicagochica5 Group Title
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    on the previous question?

    • one year ago
  16. miteshchvm Group Title
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    your question can be written as\[{ \frac{ (x^2 + 2x + 1) }{ x-2 } } \div { \frac{ x^2 - 1 }{ x^2 - 4 } } \] now \[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \] \[\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\] \]

    • one year ago
  17. miteshchvm Group Title
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    you get it?

    • one year ago
  18. chicagochica5 Group Title
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    im getting a better understand of it

    • one year ago
  19. chicagochica5 Group Title
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    A:

    • one year ago
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  20. chicagochica5 Group Title
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    B:

    • one year ago
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  21. chicagochica5 Group Title
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    C:

    • one year ago
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  22. chicagochica5 Group Title
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    D:

    • one year ago
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  23. miteshchvm Group Title
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    on eliminating values you get \[\frac{ (x+1)(x+2) }{ (x-1) }\]

    • one year ago
  24. chicagochica5 Group Title
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    thanks for everything @miteshchvm

    • one year ago
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