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wendy67 Group Title

SImplify: Help Please!

  • 2 years ago
  • 2 years ago

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  1. Godfrey_Idrobo_Libreros Group Title
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    ???

    • 2 years ago
  2. wendy67 Group Title
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    \[\frac{ 1+2i }{ \sqrt{2}+i }\]

    • 2 years ago
  3. No-data Group Title
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    Multiply both numerator and denominator by the conjugate of the denominator.I mean

    • 2 years ago
  4. No-data Group Title
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    \[\sqrt{2}-i\]

    • 2 years ago
  5. No-data Group Title
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    \[\frac{1+2i}{\sqrt{2}+i}\cdot \frac{\sqrt{2}-i}{\sqrt{2}-i}\]

    • 2 years ago
  6. wendy67 Group Title
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    I got: \[\frac{ \sqrt{2}-i+2i \sqrt{2}-2i ^{2} }{ 2-i ^{2}}\]

    • 2 years ago
  7. No-data Group Title
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    Great!

    • 2 years ago
  8. No-data Group Title
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    but also you know that \[i^2=-1\]

    • 2 years ago
  9. wendy67 Group Title
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    yes

    • 2 years ago
  10. wendy67 Group Title
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    so, it would be: \[\frac{ \sqrt{2}-i+2i \sqrt{2}-2(-1) }{ 2-(-1) }\]

    • 2 years ago
  11. No-data Group Title
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    yes, now just make arithmetic operations.

    • 2 years ago
  12. wendy67 Group Title
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    ok i got: \[\frac{ \sqrt{2}-i+2i \sqrt{2}+2 }{ 3 }\] and this where i got stuck

    • 2 years ago
  13. No-data Group Title
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    is there a specific anwer you need to get?

    • 2 years ago
  14. wendy67 Group Title
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    yes i should get : \[\frac{ 2+\sqrt{2} }{ 3 }+\frac{ 2\sqrt{2}-1 }{ 3 }\]

    • 2 years ago
  15. No-data Group Title
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    I got: \[\frac{\sqrt{2}+2}{3}-\frac{1+2\sqrt{2}}{3}i\]

    • 2 years ago
  16. wendy67 Group Title
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    ok so how did you get that?

    • 2 years ago
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