Matemática, perguntado por Usuário anônimo, 8 meses atrás

a)Qual é a Raiz de 8 x Raiz Cúbica de 4

B)Raiz cúbica de 6 divididos por Raiz quarta de n3

C) Raiz quadrada de 45a x Raiz quadrada de 2 = 15 raiz sexta de 8

D) Raiz 7 de 128 + Raiz quadrada de a⁶ (divididos) por Raiz Cúbica de a⁶ igual a Raiz cúbica de 512 - Raiz quarta de 16a⁴ divididos por Raiz quinta de 32​

Soluções para a tarefa

Respondido por Emerre
2

Radiciação

a) Qual é a Raiz de 8 x Raiz Cúbica de 4

a) \sqrt[n]{x} .\sqrt[3]{4} \\

Vamos primeiro decompor o número 8 em fatores primos

8   2

4   2

2   2

1    1

8=2³

Então ficará assim:

\sqrt{2^3}.\sqrt[3]{2^2}= \\\\

Vamos retirar do radical

2^\frac{3}{2} .2^\frac{2}{3} =

Agora temos bases iguais e expoentes diferentes,

Fazer o mmc dos expoentes = 6

4\frac{9+2}{6} =\\\\\\4\frac{11}{6} =\\\\\\\sqrt[6]{4^1^1} =\\\\\\\sqrt[6]{4^6.4^5} =\\\\\\4\sqrt[6]{4^5} =\\\\\\Resposta=4\sqrt[6]{1024}

b) Raiz cúbica de 6 divididos por Raiz quarta de 3

b)\sqrt[3]{6} :\sqrt[4]{3} =\\\\\\\dfrac{\sqrt[3]{6} }{\sqrt[4]{3} } =\\\\

Retirando do radical

\dfrac{6\frac{1}{3} }{3\frac{1}{4} } =\\\\\\2\frac{1}{3} .1^-\frac{1}{4} =\\\\\\mmc=12\\\\\\2\frac{1}{12} =\\\\\\Resposta=\sqrt[12]{2} \\\\

c) Raiz quadrada de 45a x Raiz quadrada de 2 = 15 raiz sexta de 8

\sqrt{45a} .\sqrt{2} =15\sqrt[6]{8} \\\\\\\sqrt{5.3^2.a} .2^\frac{1}{2} =15\sqrt[6]{2^3} \\\\\\3\sqrt{5a} .2^\frac{1}{2} =15.2^\frac{3}{6} \\\\\\3.5a^\frac{1}{2} .2^\frac{1}{2} =30^\frac{1}{2} \\\\\\30a^\frac{1}{2} ^+^\frac{1}{2} =30^\frac{1}{2} \\\\\\30a^\frac{2}{2} =30^\frac{1}{2} \\\\\\30a=30^\frac{1}{2} \\\\a=\dfrac{30^\frac{1}{2} }{30} \\\\\\a=1^\frac{1}{2} \\\\\\a=\sqrt{1}  \\\\\\a=1\\\\

d) Raiz 7 de 128 + Raiz quadrada de a⁶ (divididos) por Raiz Cúbica de a⁶ igual a Raiz cúbica de 512 - Raiz quarta de 16a⁴ divididos por Raiz quinta de 32

\sqrt[7]{128} +\dfrac{\sqrt{a^6} }{\sqrt[3]{a^6} } =\sqrt[3]{512} -\dfrac{\sqrt[4]{16a^4} }{\sqrt[5]{32} } \\\\\\\\\sqrt[7]{2^7} +\dfrac{\sqrt{a^2.a^2.a^2} }{\sqrt[3]{a^3.a^3} } =\sqrt[3]{2^9} -\dfrac{\sqrt[4]{2^4.a^4} }{\sqrt[5]{2^5} } \\\\\\\\2+\dfrac{a.a.a}{a.a} =\sqrt[3]{2^3.2^3.2^3} -\dfrac{2.a}{2}\\\\\\2+a=2.2.2-a\\\\\\\\a+a=8 -2\\\\\\\\2a=6\\\\\\\\a=\dfrac{6}{2} \\\\\\Resposta=a=3\\

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Anexos:
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