(√10/√5)⁻² ⁴√(√7/√14)² ∛(27/9)² (√10/5+4)° +√7.... Question from @ikersito

August 2018 1 117 Report

(√10/√5)⁻²

⁴√(√7/√14)²

∛(27/9)²

(√10/5+4)° +√7

Piscis04 $( \frac{ \sqrt{10} }{ \sqrt{5}})^{-2}= ( \frac{ \sqrt{5} }{ \sqrt{10}})^{2}\ invertimos \ la \ fracci\'on \ para\ sacar \ el \ menos \\ \\ ( \frac{ \sqrt{10} }{ \sqrt{5}})^{-2}= ( \frac{ \sqrt{5} }{ \sqrt{5}* \sqrt{2} })^{2} \\ \\ ( \frac{ \sqrt{10} }{ \sqrt{5}})^{-2}= ( \frac{ 1 }{ \sqrt{2}})^{2}\\ \\ ( \frac{ \sqrt{10} }{ \sqrt{5}})^{-2}= \boxed{ \frac{1}{2}}\qquad resolvemos \ la \ potencia \\ \\ \\$
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$\sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{7} *\sqrt{2} })^2} \\ \\ \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \sqrt[2]{( \frac{ \sqrt{7}}{ \sqrt{7} *\sqrt{2} })} \\ \\ \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \sqrt[2]{( \frac{ 1}{\sqrt{2} })} \\ \\ \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \sqrt[4]{ \frac{1}{2}} \\ \\ \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \frac{ \sqrt[4]{1}}{ \sqrt[4]{2}}\qquad \ hay \ que \ racionalizar$

$\sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \frac{ 1*\sqrt[4]{2^3}}{ \sqrt[4]{2}* \sqrt[4]{2^3} } \\ \\ \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \frac{ \sqrt[4]{8}}{ \sqrt[4]{2^4} } \\ \\ \sqrt[4]{( \frac{ \sqrt{7}}{ \sqrt{14}})^2}= \frac{ \sqrt[4]{8}}{ 2 } \\ \\$
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$\sqrt[3]{( \frac{27}{9})^2}= \sqrt[3]{(3)^2} \\ \\ \sqrt[3]{( \frac{27}{9})^2}= \boxed{ \sqrt[3]{9}}$
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$( \sqrt{ \frac{10}{5}+4})^0+ \sqrt{7}= ( \sqrt{ 2+4})^0+ \sqrt{7} \\ \\ ( \sqrt{ \frac{10}{5}+4})^0+ \sqrt{7}= ( \sqrt{ 6})^0+ \sqrt{7} \\ \\ ( \sqrt{ \frac{10}{5}+4})^0+ \sqrt{7}= \boxed{1+ \sqrt{7}}$

Espero que te sirva, salu2!!!!

3 votes Thanks 3

ikersito graciassss te adoro....

Piscis04 De nada!!!! :D