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constantinn
Pe: 2008-02-26T12:33:59+02:00 In: In:

clasa V-a

Comparati numerele:a=(4 la puterea 1003) +( 512 la puterea 223) + 1 si
b=( 32 la puterea 401) + (256 la puterea 251) + 1

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3 raspunsuri

  1. veroniquero

    hy

        \[ \begin{array}{l} a = 4^{1003} + 512^{223} + 1 = \left( {2^2 } \right)^{1003} + \left( {2^9 } \right)^{223} + 1 = \\ = 2^{2006} + 2^{2007} + 1 = 2^{2005} \cdot \left( {2^1 + 2^2 + 1} \right) = 2^{2005} \cdot 7 \\ \end{array} \]

        \[ \begin{array}{l} b = 32^{401} + 256^{251} + 1 = \left( {2^5 } \right)^{401} + \left( {2^8 } \right)^{251} + 1 = \\ = 2^{2005} + 2^{2008} + 1 = 2^{2005} \cdot \left( {1 + 2^3 + 1} \right) = 2^{2005} \cdot 10 \\ \end{array} \]

        \[ a = 2^{2005} \cdot 7 \]

        \[ b = 2^{2005} \cdot 10 \]

    deci b>a

  2. marius00

    Rezolvare incorecta
    1 nu intra la putere
    a=6*2^2005+1
    b=9*2^2005+1
    deci b>a

  4. veroniquero

    da, marius00, ai dreptate, 1 nu intra la factorul comun…

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