• Matéria: Matemática
  • Autor: caminhodepaim
  • Perguntado 6 anos atrás

calcule as operacoes com radicais:

Anexos:

Respostas

respondido por: Makaveli1996
1

Oie, Td Bom?!

c)

 = 3 \sqrt[3]{2}  - 7 \sqrt[3]{2}  - 6 \sqrt[3]{2}

 = (3 - 7 - 6) \sqrt[3]{2}

 = (3 - 13) \sqrt[3]{2}

 =  - 10 \sqrt[3]{2}

d)

 =  \sqrt{15}   \sqrt{6}

 =  \sqrt{15 \: . \: 6}

 =  \sqrt{90}

 =  \sqrt{3 {}^{2}  \: . \: 10}

 =  \sqrt{3 {}^{2} }  \sqrt{10}

 =3  \sqrt{10}

e)

 = 3 \sqrt[5]{6}  \: . \: 4 \sqrt[5]{5}

 = 12 \sqrt[5]{6}  \sqrt[5]{5}

 = 12 \sqrt[5]{6 \: . \: 5}

 = 12 \sqrt[5]{30}

f)

 =  \sqrt[3]{128}  \div  \sqrt[3]{2}

 =  \sqrt[3]{128 \div 2}

 =  \sqrt[3]{64}

 =  \sqrt[3]{4 {}^{3} }

 = 4

g)

 =  \sqrt{8}  \div  \sqrt{2}

 =  \sqrt{8 \div 2}

 =  \sqrt{4}

 =  \sqrt{2 {}^{2} }

 = 2

Att. Makaveli1996

respondido por: Anônimo
1

Explicação passo-a-passo:

c)

\sf 3\sqrt[3]{2}-7\sqrt[3]{2}-6\sqrt[3]{2}

\sf =-4\sqrt[3]{2}-6\sqrt[3]{2}

\sf =\red{-10\sqrt[3]{2}}

d)

\sf \sqrt{15}\cdot\sqrt{6}

\sf =\sqrt{90}

\sf =\sqrt{9\cdot10}

\sf =\red{3\sqrt{10}}

e)

\sf 3\sqrt[5]{6}\cdot4\sqrt[5]{5}

\sf =3\cdot4\cdot\sqrt[5]{6}\cdot\sqrt[5]{5}

\sf =\red{12\sqrt[5]{30}}

f)

\sf \sqrt[3]{128}\div\sqrt[3]{2}

\sf =\sqrt[3]{64}

\sf =\sqrt[3]{4^3}

\sf =\red{4}

g)

\sf \sqrt{8}\div\sqrt{2}

\sf =\sqrt{4}

\sf =\sqrt{2^2}

\sf =\red{2}

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