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Cel mai inteligent răspuns!
2014-10-18T14:31:19+03:00
Partea intreaga are proprietatea  [x] \leq x <[x]+1.

De aici, deducem:  
2010 \leq x <2011\\ \\ \\ \dfrac{1}{2010} \geq \dfrac{1}{x}>\dfrac{1}{2011}.

Tot din prima proprietate, se deduce:

 x-1<[x] \leq x \\ \\ \\ \dfrac{1}{x}-1<\left[\dfrac{1}{x}\right] \leq \dfrac{1}{x}\\ \\ \\ \dfrac{1}{2010}-1<\left[\dfrac{1}{x}\right] \leq \dfrac{1}{2011}

Stim ca x este pozitiv si ca \dfrac{1}{2011}<1, de unde deducem:

\left[\dfrac{1}{x}\right]=0

\left[\dfrac{1+x}{x}\right]=\left[\dfrac{1}{x}+1\right]=\left[\dfrac{1}{x}\right]+1=1.


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