Question

16) Determine as somas algébricas:
a)
$\frac{7}{3} \sqrt[3]{2} - 2 \sqrt[3]{2} - \frac{5}{4} \sqrt[3]{2} =$
b)
$5 \sqrt[3]{2} - 8 \sqrt[3]{3} + 2 - 4 \sqrt[3]{2} + 8 \sqrt[3]{3 = }$
c)
$8 \sqrt[5]{7} + \sqrt[4]{6} - 12 \sqrt[5]{7} - 10 \sqrt[4]{6} =$
d)
$\frac{ \sqrt{5} }{6} + \frac{ \sqrt{5} }{2} - \frac{ \sqrt{5} }{5} - \frac{ \sqrt{5} }{3} =$
Quem souber,me ajude...Agradeço desde já ​

Answer 1

Explicação passo-a-passo:

$a)\\ \\ {7\over3}\sqrt[3]{2} -2\sqrt[3]{2} -{5\over4}\sqrt[3]{2} =\\ \\ \sqrt[3]{2} ~~({7\over3}-2-{5\over4})=\\ \\ \sqrt[3]{2} ~({{28-24-15\over12})=\\ \\ -{11\over12}\sqrt[3]{2}$

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$b)\\ 5\sqrt[3]{2} -8\sqrt[3]{3} +2-4\sqrt[3]{2} +8\sqrt[3]{3} =\\ \\ 5\sqrt[3]{2} -4\sqrt[3]{2} -\not8\sqrt[3]{3} +\not8\sqrt[3]{3} +2=\\ \\ \sqrt[3]{2} +2$

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$c)\\ \\ 8\sqrt[5]{7} -12\sqrt[5]{7} +\sqrt[4]{6} -10\sqrt[4]{6} =\\ \\ -4\sqrt[5]{7} -9\sqrt[4]{6}$

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$mmc(6,2,5,3)=30\\ \\ \sqrt{5} ~~({1\over6}+{1\over2}-{1\over5}-{1\over3})=\\ \\ \sqrt{5} ({5+15-6-10\over30})=\\ \\ \sqrt{5} ({20-16\over30})=\\ \\ \sqrt{5} ({-4\over30})=\\ \\ -{2\over15}\sqrt{5}$