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Pe: 2010-10-07T15:15:43+03:00 In: In:

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(4la puterea n+1ORI3LAPUTEREA 2n+2 plus2la puterea2n ori9la puterean+2)impartitla6la puterea2n

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  1. GreatMath
    veteran (III)

        \[ \begin{array}{l} n = \frac{{{\rm{4}}^{{\rm{n}} + {\rm{1}}} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{2 }}} {\rm{ + 2}}^{{\rm{2n}}} \cdot {\rm{9}}^{{\rm{n}} + {\rm{2}}} }}{{{\rm{6}}^{{\rm{2n}}} }} = \frac{{\left( {{\rm{2}}^2 } \right)^{{\rm{n}} + {\rm{1}}} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{2 }}} {\rm{ + 2}}^{{\rm{2n}}} \cdot \left( {{\rm{3}}^2 } \right)^{{\rm{n}} + {\rm{2}}} }}{{{\rm{6}}^{{\rm{2n}}} }} \\ = \frac{{{\rm{2}}^{{\rm{2n}} + 2} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{2 }}} {\rm{ + 2}}^{{\rm{2n}}} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{4}}} }}{{{\rm{6}}^{{\rm{2n}}} }} = \frac{{{\rm{4}} \cdot {\rm{2}}^{{\rm{2n}}} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{2 }}} {\rm{ + 9}} \cdot {\rm{2}}^{{\rm{2n}}} \cdot {\rm{3}}^{{\rm{2n}} + 2} }}{{{\rm{6}}^{{\rm{2n}}} }} \\ = \frac{{{\rm{2}}^{{\rm{2n}}} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{2 }}} \left( {{\rm{4 + 9}}} \right)}}{{{\rm{6}}^{{\rm{2n}}} }} = \frac{{{\rm{13}} \cdot {\rm{2}}^{{\rm{2n}}} \cdot {\rm{3}}^{{\rm{2n}} + {\rm{2 }}} }}{{{\rm{6}}^{{\rm{2n}}} }} = \frac{{{\rm{13}} \cdot 9\left( {{\rm{2}}^{{\rm{2n}}} \cdot {\rm{3}}^{{\rm{2n }}} } \right)}}{{{\rm{6}}^{{\rm{2n}}} }} = \frac{{{\rm{13}} \cdot 9 \cdot 6^{2n} }}{{{\rm{6}}^{{\rm{2n}}} }} = {\rm{13}} \cdot 9 \\ \end{array} \]

    Urmareste solutia pas cu pas.
    Pentru a intelege deplin solutia prezentata iti recomand sa citesti:
    http://pasionatidematematica.blogspot.com/2010/08/ridicarea-la-putere-lectia-1.html , http://pasionatidematematica.blogspot.com/2010/08/puterea-unui-numar-lectia-2.html sunt sigur ca ti vor fi de ajutor.

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