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[Risolto] Proprietà potenze

  

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@vj 👍👌👍



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1024 = (2^4)^2*2^2 = 2^(8+2) = 2^10

 (2^10)^(1/5) = 2^(10/5) = 2^2 = 4 

 



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625 = 25*25 = 5^2*5^2 = 5^4

(5^4)^(1/3) = 5^(4/3)



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121^2 = 11^2*11^2 = 11^(2+2) = 11^4

(121^2)^(1/3) = (11^4)^1/3 = 11^(4/3)



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-9^2 = 3^2*3^2 = 3^4

(3^4)^(1/7)  = 3^(4/7)



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x^(11/7) * y^(2/7) 



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121 = 11^2

(11)^(2/6) = 11^(1/3) 



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======================================================

K) → $ \sqrt[5]{1024} = \sqrt[5]{2^{10}} = 2^{\frac{\cancel{10}^2}{\cancel5_1}} = 2^2 = 4;$

 

L) → $ \sqrt[3]{625}  =\sqrt[3]{5^4} = 5^{\frac{4}{3}};$

 

M) → $ \sqrt[3]{121^2} = \sqrt[3]{11^{2×2}} = \sqrt[3]{11^4} = 11^{\frac{4}{3}};$

 

N) → $ \sqrt[7]{-(9)^2} = -(9)^{\frac{2}{7}}.$

@gramor 👍👌👍

@remanzini_rinaldo - Grazie mille, buona giornata.



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