Como se faz para simplificar os radicais? a) √3elevado a 5 . 2elevado a 4. 7²b) ∛5 elevado a 7 . 6² .... Pergunta de ideia deanadelmiro

October 2020 1 80 Report

Como se faz para simplificar os radicais?
a) √3elevado a 5 . 2elevado a 4. 7²

b) ∛5 elevado a 7 . 6² . 7³

c) ∛1296

d) raiz quinta√ 15552

Obrigada.

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Niiya

Verified answer

A)

$\sqrt{3^{5}*2^{4}*7^{2}}=\sqrt{3^{4}*3^{1}*2^{4}*7^{2}}\\\sqrt{3^{5}*2^{4}*7^{2}}=\sqrt{3^{4}*2^{4}*7^{2}*3}\\\sqrt{3^{5}*2^{4}*7^{2}}=\sqrt{3^{4}}*\sqrt{2^{4}}*\sqrt{7^{2}}*\sqrt{3}}$
$\sqrt{3^{5}*2^{4}*7^{2}}=\sqrt[2]{3^{4}}*\sqrt[2]{2^{4}}*\sqrt[2]{7^{2}}*\sqrt{3}\\\sqrt{3^{5}*2^{4}*7^{2}}=3^{(4/2)}*2^{(4/2)}*7^{(2/2)}*\sqrt{3}\\\sqrt{3^{5}*2^{4}*7^{2}}=3^{2}*2^{2}*7^{1}*\sqrt{3}\\\sqrt{3^{5}*2^{4}*7^{2}}=9*4*7*\sqrt{3}\\\sqrt{3^{5}*2^{4}*7^{2}}=252\sqrt{3}$

b)

$\sqrt[3]{5^{7}*6^{2}*7^{3}}=\sqrt[3]{5^{6}*5^{1}*6^{2}*7^{3}}\\\sqrt[3]{5^{7}*6^{2}*7^{3}}=\sqrt[3]{5^{6}*7^{3}*6^{2}*5}\\\sqrt[3]{5^{7}*6^{2}*7^{3}}=\sqrt[3]{5^{6}}*\sqrt[3]{7^{3}}*\sqrt[3]{6^{2}*5}\\\sqrt[3]{5^{7}*6^{2}*7^{3}}=5^{(6/3)}*7^{(3/3)}*\sqrt[3]{36*5}\\\sqrt[3]{5^{7}*6^{2}*7^{3}}=5^{2}*7^{1}*\sqrt[3]{180}\\\sqrt[3]{5^{7}*6^{2}*7^{3}}=25*7*\sqrt[3]{180}\\\sqrt[3]{5^{7}*6^{2}*7^{3}}=175\sqrt[3]{180}$

c)

1296 | 3
0432 | 3
0144 | 3
0048 | 3
0016 | 4
0004 | 4
0001

$1296=3*3*3*3*4*4=3^{4}*4^{2}$

$\sqrt[3]{1296}=\sqrt[3]{3^{4}*4^{2}}\\\sqrt[3]{1296}=\sqrt[3]{3^{4}*(2^{2})^{2}}\\\sqrt[3]{1296}=\sqrt[3]{3^{3}*3^{1}*2^{(2*2)}}\\\sqrt[3]{1296}=\sqrt[3]{3^{3}*2^{4}*3}\\\sqrt[3]{1296}=\sqrt[3]{3^{3}*2^{3}*2^{1}*3}\\\sqrt[3]{1296}=\sqrt[3]{3^{3}*2^{3}*2*3}\\\sqrt[3]{1296}=\sqrt[3]{3^{3}}*\sqrt[3]{2^{3}}*\sqrt[3]{2*3}\\\sqrt[3]{1296}=3^{(3/3)}*2^{(3/3)}*\sqrt[3]{6}\\\sqrt[3]{1296}=3^{1}*2^{1}*\sqrt[3]{6}\\\sqrt[3]{1296}=3*2*\sqrt[3]{6}\\\sqrt[3]{1296}=6\sqrt[3]{6}$

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anadelmiro Te amo!!!!!!!!

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