Question

Podrian ayudarme a resolver, si pueden explicarme tambien, se los agradeceria.

Answer 1

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Answer 2

1)  
$5 \sqrt{8} -3 \sqrt{2} + \sqrt{5} +5 \sqrt{32} -4 \sqrt{5} = \\ \\ =5 \sqrt{2^3} -3 \sqrt{2} + \sqrt{5} +5 \sqrt{2^5} -4 \sqrt{5}= \\ \\ =5*2 \sqrt{2} -3 \sqrt{2} + \sqrt{5} +5*2^2 \sqrt{2} -4 \sqrt{5} = \\ \\= 10 \sqrt{2} -3 \sqrt{2} + \sqrt{5} +20 \sqrt{2} -4 \sqrt{5} = \\ \\ =27 \sqrt{2} -3 \sqrt{5} = \\ \\ =3*(9 \sqrt{2}- \sqrt{5})$
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3) 
$\sqrt{2x^2y^4}-3xy^2 \sqrt{2}+5 \sqrt{8x^2y^4}= \\ \\ =xy^2 \sqrt{2}-3xy^2 \sqrt{2}+5*2xy^2 \sqrt{2}= \\ \\ =8xy^2 \sqrt{2}$
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4) 
$\sqrt[3]{16} + \sqrt[3]{2} - \sqrt[3]{128} = \\ \\ =\sqrt[3]{2^4} + \sqrt[3]{2} - \sqrt[3]{2^7}= \\ \\ = \sqrt[3]{2^3*2^1} +\sqrt[3]{2} - \sqrt[3]{2^3*2^3*2} = \\ \\ =2 \sqrt[3]{2} +\sqrt[3]{2} - 4 \sqrt[3]{2} = \\ \\ =- \sqrt[3]{2}$
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6)  
$\sqrt{ \frac{72}{2} } = \frac{ \sqrt{72} }{ \sqrt{2} } = \frac{\sqrt{2^3*3^2} }{ \sqrt{2} } = \frac{2*3* \sqrt{2} }{ \sqrt{2} } =6$
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Saludos.