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ksaimouli

  • 3 years ago

y^4+(1/16y^4)+1/2

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  1. ksaimouli
    • 3 years ago
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    \[y^4+\frac{ 1 }{ 16y^4 }+\frac{ 1 }{ 2 }\]

  2. ksaimouli
    • 3 years ago
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    perfect square

  3. ksaimouli
    • 3 years ago
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    @Jonask

  4. phi
    • 3 years ago
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    you could make 16y^4 the common denominator and add up the terms the denominator is a perfect square. Does the numerator factor?

  5. ksaimouli
    • 3 years ago
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    is their any easy way instead of just taking common denominator

  6. ksaimouli
    • 3 years ago
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    @zepdrix

  7. ksaimouli
    • 3 years ago
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    i know a^2+2ab+b^2

  8. phi
    • 3 years ago
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    It is not that hard

  9. phi
    • 3 years ago
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    16 y^8 + 1 + 8 y^4

  10. ksaimouli
    • 3 years ago
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    =(a+b)^2

  11. phi
    • 3 years ago
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    re-arrange as 16 y^8 + 8 y^4 + 1 if you let y^4 = x, this is also 16x^2 + 8x +1

  12. Jonask
    • 3 years ago
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    i just tried\[(y^2+\frac{1}{4y^2})^2\]

  13. phi
    • 3 years ago
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    if it is square your only hope is (4x+1)^2 which works

  14. ksaimouli
    • 3 years ago
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    is that =0 because what happened to common denominator

  15. ksaimouli
    • 3 years ago
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    (4x^4+1)^2

  16. Jonask
    • 3 years ago
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    is it not true to say\[(y^2+\frac{1}{4y^2})^2=y^4+2(y^2)(\frac{1}{4y^2})+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}\]

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