Question

Ayudenme porfavor necesito esto...
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Answer 1

Opción A.

$\sqrt[3]{4}\cdot 4^{\frac{1}{3}}+\sqrt[5]{4}\cdot 4^{\frac{3}{5}} = 4^{\frac{1}{3}}\cdot \:4^{\frac{1}{3}}+4^{\frac{1}{5}}\cdot \:4^{\frac{3}{5}} = 4^{\frac{2}{3}}+4^{\frac{1}{5}}\cdot \:4^{\frac{3}{5}} =4^{\frac{2}{3}}+4^{\frac{4}{5}}$

Opción B.

$2\sqrt{8}+3\sqrt{18}-2\sqrt{16}+5\sqrt{32} =$

$2\sqrt{8}=4\sqrt{2}$

$3\sqrt{18}=9\sqrt{2}$

$2\sqrt{16}=8$

$5\sqrt{32}=20\sqrt{2}$

$=4\sqrt{2}+9\sqrt{2}-8+20\sqrt{2} =33\sqrt{2}-8$

Opción C.

$\sqrt[3]{3}+2\sqrt[3]{24}-5\sqrt[3]{81}-\sqrt[3]{64}=$

$2\sqrt[3]{24}=4\sqrt[3]{3}$

$5\sqrt[3]{81}=15\sqrt[3]{3}$

$\sqrt[3]{64}=4$

$=\sqrt[3]{3}+4\sqrt[3]{3}-15\sqrt[3]{3}-4 = -10\sqrt[3]{3}-4$

Opción D.

$6+5\sqrt{27}-\left(-15\right)+2\sqrt{8}+4\sqrt{12}=6+5\sqrt{27}+15+2\sqrt{8}+4\sqrt{12}$

$5\sqrt{27}=15\sqrt{3}$

$2\sqrt{8}=4\sqrt{2}$

$4\sqrt{12}=8\sqrt{3}$

$4\sqrt{12}=8\sqrt{3}$

$=6+15\sqrt{3}+15+4\sqrt{2}+8\sqrt{3}=6+23\sqrt{3}+15+4\sqrt{2}=21+23\sqrt{3}+4\sqrt{2}$

Opción E.

$2+4\sqrt{8}-\left(-30\right)+2\sqrt{8}+4\sqrt{18}$

$=2+4\sqrt{8}+30+2\sqrt{8}+4\sqrt{18} =$

$6\sqrt{8}=12\sqrt{2}$

$4\sqrt{18}=12\sqrt{2}$

$=2+12\sqrt{2}+30+12\sqrt{2}$

$=2+24\sqrt{2}+30$

$=32+24\sqrt{2}$