Question

Me ajudem a resolver duas expressões? A=8^1/3+(1/9)^1/2+16^1/4 e A=(0,25)^0,5+18^0,25+1,6^-0,5 Me expliquem por favor, obrigado desde já!!

Answer 1

Olá, Aprendiz.

 

$<var>A=8^{\frac13}+(\frac19)^{\frac12}+16^{\frac14}=\sqrt[3]{2^3}+\sqrt\frac1{3^2}+\sqrt[4]{2^4}=2+\frac13+2=\frac{13}3</var>$

 

$<var>A=(0,25)^{0,5}+18^{0,25}+1,6^{-0,5}=(\frac14)^{\frac12}+18^{\frac14}+(\frac{16}{10})^{-\frac12}=\\ =\sqrt\frac1{2^2}+\sqrt[4]{18}+\sqrt{(\frac45)^{-1}}=\frac12+\sqrt[4]{18}+\sqrt{\frac54}=\frac12+\sqrt[4]{18}+\frac{\sqrt5}2=\\ =\sqrt[4]{18}+\frac{1+\sqrt5}2</var>$

Answer 2

acredito que no segundo era 81 e não 18 !!

Vamos lá:

 

$<var>A = \sqrt[3]{8} +\sqrt[2]{\frac{1}{9}} + \sqrt[4]{16} \\ A = \sqrt[3]{2^{3}} +\sqrt[2]{(\frac{1}{3})^{2}} + \sqrt[4]{2^{4}} \\ A = 2 + \frac{1}{3} +2 \\ \ fazendo\ mmc = 3 \\ A = \frac{3.2}{3} +\frac{1}{3}+\frac{3.2}{3}\\ \\A = \frac{6}{3} +\frac{1}{3}+\frac{6}{3}\\ \\ A = \frac{13}{3}\\ \\ \\ \\ A = \sqrt[2]{0,25} +\sqrt[4]{81} + \sqrt[2]{1,6^{-1}} \\ \\ A = \sqrt[2]{\frac{1}{4}} +\sqrt[4]{81} + \sqrt[2]{1,6^{-1}} \\ \\ A = \sqrt[2]{\frac{1}{4}} +\sqrt[4]{81} + \sqrt[2]{\frac{16}{10}} ^{-1}} \\ \\ </var>$

 

A = 1/2 + 4 +V10/4

A = 2/4+ 16/4+ V10/4

A = (18+V10)/4