Question

Efetue as operações





Simplifique as expressões

Answer 1

Explicação passo-a-passo:

1- a) $\sqrt{7}-5\sqrt{7}+2\sqrt{7}=(1-5+2)\sqrt{7}=-2\sqrt{7}$

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   b) $2\sqrt[5]{2}+\sqrt[5]{64}$

       Fatorando o 64, temos: $64=2^{6}=2^{5}.2$

       $2\sqrt[5]{2}+\sqrt[5]{2^{5}.2}=2\sqrt[5]{2}+2\sqrt[5]{2}=(2+2)\sqrt[5]{2}=4\sqrt[5]{2}$

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   c) $2\sqrt{16}+3\sqrt[3]{16}+4\sqrt[4]{16}+\sqrt[6]{16}$

       Fatorando o 16, dependendo do índice do radical, fica

       $\sqrt{16}=\sqrt{4^{2}}$ ; $\sqrt[3]{16}=\sqrt[3]{2^{3}.2 }$ ; $\sqrt[4]{16}=\sqrt[4]{2^{4}}$ ; $\sqrt[6]{16}=\sqrt[3]{\sqrt{4^{2}}}$

       Substituindo fica

       $2\sqrt{4^{2}}+3\sqrt[3]{2^{3}.2}+4\sqrt[4]{2^{4}}+\sqrt[3]{\sqrt{4^{2}}}=$

       $8+3.2\sqrt[3]{2}+4.2+\sqrt[3]{4}=8+6\sqrt[3]{2}+8+\sqrt[3]{4}=16+6\sqrt[3]{2}+\sqrt[3]{4}$

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   d) $(3\sqrt{2}+7\sqrt{3})+(6\sqrt{2}-2\sqrt{3})$

       Eliminando os parênteses fica

       $3\sqrt{2}+7\sqrt{3}+6\sqrt{2}-2\sqrt{3}=(3+6)\sqrt{2}+(7-2)\sqrt{3}=9\sqrt{2}+5\sqrt{3}$

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   e) $\sqrt{32}-2\sqrt{12}-\sqrt{75}+3\sqrt{72}$

       Fatorando o 32, 12, 75 e 72 fica

                  $32=2^{5}$ ; $12=2^{2}.3$ ; $75=5^{2}.3$ ; $72=2^{3}.3^{2}$

       Substituindo fica

       $\sqrt{2^{5}}-2\sqrt{2^{2}.3}-\sqrt{5^{2}.3}+3\sqrt{2^{3}.3^{2}}=$

       $\sqrt{2^{2}.2^{2}.2}-2.2\sqrt{3}-5\sqrt{3}+3.3\sqrt{2^{2}.2}=$

       $2.2\sqrt{2}-4\sqrt{3}-5\sqrt{3}+9.2\sqrt{2}=$

       $4\sqrt{2}-4\sqrt{3}-5\sqrt{3}+18\sqrt{2}=(4+18)\sqrt{2}+(-4-5)\sqrt{3}=22\sqrt{2}-9\sqrt{3}$

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2- a) $3\sqrt[3]{2}-7\sqrt[3]{2}-6\sqrt[3]{2}=(3-7-6)\sqrt[3]{2}=-10\sqrt[3]{2}$

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   b) $\sqrt{12}-\sqrt[4]{9}-\sqrt[6]{27}-\sqrt[8]{81}$

       Fatorando o 12, 9, 27 e 81 fica

             $12=2^{2}.3$ ; $9=3^{2}$ ; $27=3^{3}$ ; $81=3^{4}$

       Substituindo fica

       $\sqrt{2^{2}.3}-\sqrt[4]{3^{2}}-\sqrt[6]{3^{3}}-\sqrt[8]{3^{4}}=$

       $2\sqrt{3}-\sqrt{\sqrt{3^{2}}}-\sqrt{\sqrt[3]{3^{3}}}-\sqrt{\sqrt[4]{3^{4}}}=$

       $2\sqrt{3}-\sqrt{3}-\sqrt{3}-\sqrt{3}=(2-1-1-1)\sqrt{3}=-\sqrt{3}$

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   c) $2\sqrt{150}-4\sqrt{54}+6\sqrt{24}$

       Fatorando o 150, 54 e 24 fica

           $150=2.3.5^{2}$ ; $54=2.3^{3}=2.3^{2}.3$ ; $24=2^{3}.3=2^{2}.2.3$

      Substituindo fica

      $2\sqrt{2.3.5^{2}}-4\sqrt{2.3^{2}.3}+6\sqrt{2^{2}.2.3}=$

      $2.5\sqrt{6}-4.3\sqrt{6}+6.2\sqrt{6}=10\sqrt{6}-12\sqrt{6}+12\sqrt{6}=(10-12+12)\sqrt{6}=10\sqrt{6}$

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   d) $7\sqrt{32}-5\sqrt{2}+\sqrt{8}$

       Fatorando o 32 e 8 fica

            $32=2^{5}=2^{2}.2^{2}.2$ ; $8=2^{3}=2^{2}.2$

       Substituindo fica

       $7\sqrt{2^{2}.2^{2}.2}-5\sqrt{2}+\sqrt{2^{2}.2}=$

       $7.2.2\sqrt{2}-5\sqrt{2}+2\sqrt{2}=28\sqrt{2}-5\sqrt{2}+2\sqrt{2}=(28-5+2)\sqrt{2}=25\sqrt{2}$