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Jonask

  • 3 years ago

which one is larger

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  1. Jonask
    • 3 years ago
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    \[\sqrt[3]{60}\] \[2+\sqrt[3]{7}\]no machines allowed

  2. Jonask
    • 3 years ago
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    \[2=\sqrt[3]{8}\]

  3. Jonask
    • 3 years ago
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    \[60=(\frac{ 7+8 }{ 2 })10\]

  4. Jonask
    • 3 years ago
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    \[\sqrt[3]{60}= \sqrt[3]{5}(\sqrt[3]{7+8})\]

  5. Jonask
    • 3 years ago
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    almost simlar now to\[\sqrt[3]{8}+\sqrt[3]{7}\] HALT

  6. Jonask
    • 3 years ago
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    !!!!!!

  7. Jonask
    • 3 years ago
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    is comparing 60 with \[(2+\sqrt[3]{7})^3\]the same idea(tautology)

  8. Jonask
    • 3 years ago
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    i think we have to use inequalities \[(2+\sqrt[3]{7})^3=8+3(4)\sqrt[3]{7}+3(2)(\sqrt[3]{7})^2+7\]

  9. Jonask
    • 3 years ago
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    \[\sqrt[3]{1}<\sqrt[3]{7}<\sqrt[3]{8}\]

  10. Jonask
    • 3 years ago
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    \[15+12a+6a^2\] and 60

  11. klimenkov
    • 3 years ago
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    Interesting question!

  12. Jonask
    • 3 years ago
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    \[a=\sqrt[3]{7}\]

  13. Jonask
    • 3 years ago
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    we can allow machines

  14. Jonask
    • 3 years ago
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    \[a<2\]

  15. Jonask
    • 3 years ago
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    \[\frac{ x+y+z }{ 3 } \leq \sqrt[3]{xyz}\] useful ineq

  16. Jonask
    • 3 years ago
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    geometric mean

  17. Jonask
    • 3 years ago
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    exhausted

  18. klimenkov
    • 3 years ago
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    Try to do like this. How to evaluate \(\sqrt[3]{7}\). \(1<{7}<8 \Rightarrow 1<\sqrt[3]{7}<2\). As 7 is closer to 8 than to 1, I hope \(\sqrt[3]{7}\) is closer to 2. Lets put \(\sqrt[3]{7}=2+\alpha\). Then \(7=8+12\alpha+6\alpha^2+\alpha^3\). Since \(\alpha <1\) we can neglect \(\alpha^2 ,\alpha^3\). So, \(\alpha=-\frac1{12}\). So the approximation for \(\sqrt[3]{7}\) is \(2-\frac1{12}=\frac{23}{12}\). Do this once again, because it is not a good accuracy for this problem.

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