• Matéria: Matemática
  • Autor: laisbrr
  • Perguntado 5 anos atrás

ME AJUDE POR FAVOR
Calcule:

a) ³√1728 

b) (2√6)x(√24) 

c) (3/4)³ 

d) (√3)⁴ 

F)(√45)-(√20) 

e - tá na foto
g - tá na foto
h - tá na foto​

Anexos:

Respostas

respondido por: Anônimo
0

Explicação passo-a-passo:

a) Decompondo esse número:

\sf 1728~~~~~|~~~2

\sf ~864~~~~~~|~~~2

\sf ~432~~~~~~|~~~2

\sf ~216~~~~~~|~~~2

\sf ~108~~~~~~|~~~2

\sf ~~54~~~~~~~|~~~2

\sf ~~27~~~~~~~|~~~3

\sf ~~~9~~~~~~~~|~~~3

\sf ~~~3~~~~~~~~|~~~3

\sf ~~~1

\sf 1728=2^6\cdot3^3

Assim:

\sf \sqrt[3]{1728}=\sqrt[3]{2^6\cdot3^3}

\sf \sqrt[3]{1728}=\sqrt[3]{(2^2)^3\cdot3^3}

\sf \sqrt[3]{1728}=2^2\cdot3

\sf \sqrt[3]{1728}=4\cdot3

\sf \red{\sqrt[3]{1728}=12}

b)

\sf 2\sqrt{6}\times\sqrt{24}

\sf =2\cdot\sqrt{6\times24}

\sf =2\cdot\sqrt{144}

\sf =2\cdot12

\sf =\red{24}

c)

\sf \Big(\dfrac{3}{4}\Big)^3=\dfrac{3^3}{4^3}

\sf \Big(\dfrac{3}{4}\Big)^3=\dfrac{3\times3\times3}{4\times4\times4}

\sf \red{\Big(\dfrac{3}{4}\Big)^3=\dfrac{27}{64}}

d)

\sf (\sqrt{3})^4

\sf =\sqrt{3^4}

\sf =\sqrt{(3^2)^2}

\sf =3^2

\sf =3\times3

\sf =\red{9}

e) Lembre-se que \sf a^{\frac{b}{c}}=\sqrt[c]{a^b}

\sf (-64)^{\frac{1}{3}}

\sf =\sqrt[3]{(-64)^1}

\sf =\sqrt[3]{-64}

\sf =\sqrt[3]{-2^6}

\sf =\sqrt[3]{(-2^2)^3}

\sf =-2^2

\sf =-2\cdot2

\sf =\red{-4}

f)

\sf \sqrt{45}-\sqrt{20}

\sf =\sqrt{3^2\cdot5}-\sqrt{2^2\cdot5}

\sf =3\sqrt{5}-2\sqrt{5}

\sf =\red{\sqrt{5}}

g) Em multiplicação de potências de mesma base, repetimos a base e somamos os expoentes

\sf (-3)^5\times(-3)^{-1}

\sf =(-3)^{5+(-1)}

\sf =(-3)^{5-1}

\sf =(-3)^{4}

\sf =(-3)\times(-3)\times(-3)\times(-3)

\sf =(+9)\times(+9)

\sf =\red{+81}

h)

\sf \Big(\dfrac{\sqrt{24}}{\sqrt{2}}\Big)+3^{\frac{3}{2}}

\sf =\dfrac{\sqrt{2}\cdot\sqrt{12}}{\sqrt{2}}+\sqrt[2]{3^3}

\sf =\sqrt{12}+\sqrt{3^3}

\sf =\sqrt{2^2\cdot3}+\sqrt{3^2\cdot3}

\sf =2\sqrt{3}+3\sqrt{3}

\sf =\red{5\sqrt{3}}

respondido por: AlexandreBruckmann
0

A resposta é 5

5 \sqrt{?}3

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