Question

por favor preciso de ajudá com essas contas​

Answer 1

Explicação passo-a-passo:

$a)$  $3\sqrt{5}+\sqrt{5}-6\sqrt{5}$

   $(3+1-6)\sqrt{5}=-2\sqrt{5}$

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$b)$  $5\sqrt[5]{3}+2\sqrt[5]{3}-2\sqrt[5]{2}+1\sqrt[5]{3}$

   $(5+2-2+1)\sqrt[5]{3}=6\sqrt[5]{3}$

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$c)$  $\sqrt[3]{5}+2\sqrt[3]{5}$

   $(1+2)\sqrt[3]{5}=3\sqrt[3]{5}$

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$e)$  $\sqrt{50}+\sqrt{18}-\sqrt{8}$

   fatore o 50, 18 e 8

        50 = 2 · 5²

        18 = 2 · 3²

        8 = 2³ = 2 · 2²

   substitua

   $\sqrt{2.5^{2}}+\sqrt{2.3^{2}}-\sqrt{2.2^{2}}\\5\sqrt{2}+3\sqrt{2}-2\sqrt{2}\\(5+3-2)\sqrt{2}=6\sqrt{2}$

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$f)$  $2\sqrt{27}-5\sqrt{12}$

   fatore o 27 e 12

        27 = 3³ = 3 · 3²

        12 = 2² · 3

   substitua

   $2\sqrt{3.3^{2}}-5\sqrt{2^{2}.3}\\2.3\sqrt{3}-5.2\sqrt{3}\\6\sqrt{3}-10\sqrt{3}\\(6-10)\sqrt{3}=-4\sqrt{3}$

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$g)$  $4\sqrt{63}-\sqrt{7}$

   fatore o 63

        63 = 3² · 7

   substitua

   $4\sqrt{3^{2}.7}-\sqrt{7}\\4.3\sqrt{7}-\sqrt{7}\\12\sqrt{7}-\sqrt{7}\\(12-1)\sqrt{7}=11\sqrt{7}$

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$h)$  $\sqrt{12}+\sqrt{75}+\sqrt{108}$

   fatore o 12, 75 e 108

        12 = 2² · 3

        75 = 3 · 5²

        108 = 2² · 3³ = 2² · 3² · 3

   substitua

   $\sqrt{2^{2}.3}+\sqrt{3.5^{2}}+\sqrt{2^{2}.3^{2}.3}\\2\sqrt{3}+5\sqrt{3}+2.3\sqrt{3}\\2\sqrt{3}+5\sqrt{3}+6\sqrt{3}\\(2+5+6)\sqrt{3}=13\sqrt{3}$