Home/ Intrebari/Q 77452
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

catalina17
Pe: 2012-07-08T06:59:39+03:00 In: In:

ex puteri

Rezolvati ecuatia:

2^2460 : {(2^x+1 * 2^x+2 * 2^x+3 * … * 2^x+80) : 32^320 +

[(2^107)^6 – 2^2 * 4^4 * 8^8 * 16^16 * 32^32 * 64^64]} = (2^5^3)^4

Am facut asa:

2^2460:{(2^x*2*2^x*2^2*2^x*2^3*…*2^x*2^80):(2^5)^320 +

[2^642 – 2^2*2^8*2^64*2^160*2^384]} = 2^500

2^2460:[{2^x(2*2^2*2^3*…*2^80) ]=2^500

2^2460:[(2^x * 2^3240) : 2^1600 = 2^500

Mai departe m-am blocat nu mai stiu ce sa fac, va rog sa-mi explicati cum se face mai departe si daca este corect rezolvat!

Multumesc mult!

Similare

1 raspuns

  1. edy8
    maestru (V)

    catalina17 wrote: Rezolvati ecuatia:

    2^2460 : {(2^x+1 * 2^x+2 * 2^x+3 * … * 2^x+80) : 32^320 +

    [(2^107)^6 – 2^2 * 4^4 * 8^8 * 16^16 * 32^32 * 64^64]} = (2^5^3)^4

    2^x+1 = 2^(x+1)

    \it{\Large Se abordeaza pe fragmente (detalii) :\\\;\\(2^{x+1}\cdot 2^{x+2 }\cdot 2^{x+3 }\cdot ... \cdot 2^{x+80}) : 32^{320} = 2^{x+1+x+2+x+3+ ... +x+80} : (2^5)^{320} = \\\;\\=2^{80x + \frac{80\cdot81}{2}} : 2^{5\cdot320} = 2^{80x + 3240} : 2^{1600} = 2^{80x+3240-1600} = 2^{80x +1640} (1) \\\;\\(2^{107})^6 - 2^2 \cdot 4^4\cdot 8^8 \cdot 16^{16}\cdot 32^{32} \cdot 64^{64} = 0 (2)\\\;\\(1), (2) \Rightarrow ecuatia devine :\\\;\\2^{2460} : (2^{80x + 1640} + 0) = (2^{5^3})^4 \Leftrightarrow 2^{2460} : 2^{80x+1640} = (2^{125})^4 \Leftrightarrow\\\;\\\Leftrightarrow 2^{2460-80x-1640 } = 2^{500} \Leftrightarrow 2^{820-80x} = 2^{500} \Leftrightarrow 820-80x = 500 \Leftrightarrow\\\;\\\Leftrightarrow 820 - 500 = 80x \Leftrightarrow 320 = 80x \Leftrightarrow x = 4.\bl }

Raspunde

Raspunde