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Ralumus2007
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Media geometrica a numerelor 12∛3, 288 si 108∛9.

Răspuns :

Alexutzuu10
Media geometrica = [tex] \sqrt[3]{12 \sqrt[3]{3}*288*108 \sqrt[3]{9}} = \sqrt[3]{12*(2*12^{2})*(12*9)* \sqrt[3]{27} } = [/tex]

[tex]= \sqrt[3]{12^{4}*2*3^{2}*3} = 12 \sqrt[3]{2^{2}*3*2*3^{3}} = 36 \sqrt[3]{2^{3}*3} = 72 \sqrt[3]{3} [/tex]

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