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Mateigeorgiana7
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(1supra2+1supra3+1supra6)la puterea2011÷(4supra12)la puterea2

Răspuns :

Mariadrew

[tex]( \frac{1}{2} + \frac{1}{3} + \frac{1}{6} ) {}^{2011} \div ( \frac{4}{12}) {}^{2} \\ 1 {}^{2011} \div ( \frac{1}{3} ) {}^{2} \\ 1 \div \frac{1}{9} \\ 9[/tex]

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