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chicagochica5

  • 2 years ago

Divide.

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  1. chicagochica5
    • 2 years ago
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  2. miteshchvm
    • 2 years ago
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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

  3. chicagochica5
    • 2 years ago
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    The first part would be (x -2) 2x? + 1

  4. theredhead1617
    • 2 years ago
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    Yes!

  5. chicagochica5
    • 2 years ago
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    @theredhead1617 do you know how to solve the problem

  6. miteshchvm
    • 2 years ago
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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?

  7. chicagochica5
    • 2 years ago
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    No @miteshchvm I dont understand how to. I wish to know where to begin but i do not

  8. miteshchvm
    • 2 years ago
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    can you factorise x^2 + 2x + 1?

  9. miteshchvm
    • 2 years ago
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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?

  10. chicagochica5
    • 2 years ago
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    (x + 1) to second power?

  11. miteshchvm
    • 2 years ago
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    yes

  12. miteshchvm
    • 2 years ago
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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,

  13. chicagochica5
    • 2 years ago
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    just a little bit, its hard

  14. miteshchvm
    • 2 years ago
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    now refer to my first comment

  15. chicagochica5
    • 2 years ago
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    on the previous question?

  16. miteshchvm
    • 2 years ago
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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 \[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \] \[\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\] \]

  17. miteshchvm
    • 2 years ago
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    you get it?

  18. chicagochica5
    • 2 years ago
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    im getting a better understand of it

  19. chicagochica5
    • 2 years ago
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    A:

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  20. chicagochica5
    • 2 years ago
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    B:

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  21. chicagochica5
    • 2 years ago
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    C:

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  22. chicagochica5
    • 2 years ago
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    D:

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

  24. chicagochica5
    • 2 years ago
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    thanks for everything @miteshchvm

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