Un exercitiu usurel

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Natalee
Euclid
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Un exercitiu usurel

Post by Natalee »

Sa se determine \( k \ \in \ N \), stiind ca are loc egalitatea:

\( \frac{2^{^{3k + 5}} \ + \ 2^{^{3k+2}} \ - \ 2^{^{3k+4}}}{2^{^{2k+5}} \ - \ 2^{^{2k+3}} \ + \ 2^{^{2k+4}}} \ = \ 512 \)
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Marius Dragoi
Thales
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Post by Marius Dragoi »

\( \frac {2^{3k+2}(8+1-4)}{2^{2k+3}(4-1+2)} \) \( = 512 \) \( \Rightarrow \frac {5* 2^{3k+2}}{5* 2^{2k+3}} \) \( = 512 \) \( \Rightarrow 2^{k-1} = 2^9 \Rightarrow k=10 \) :P
Politehnica University of Bucharest
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