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Braila 2008
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Braila 2008
Posted:
Fri Nov 28, 2008 11:31 pm
by
alex2008
Demonstrati ca
\( \sqrt{\frac{a^2+2008^2}{3}} \in \mathb{R} \)
\
\( \mathb{Q} \)
\( (\forall)a\in\mathb{Z}. \)
... acolo e R minus Q ...
Posted:
Fri Nov 28, 2008 11:48 pm
by
Marius Mainea
\( a^2+2008^2=\mathcal{M}3+1 \)
sau
\( a^2+2008^2=\mathcal{M}3+2 \)