Question

De maneira semelhante. calcule (√6+√32)•(√2-√24)

Answer 1

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Calcular  $\mathsf{\big(\sqrt{6}+\sqrt{32}\big)\cdot \big(\sqrt{2}-\sqrt{24}\big):}$


Primeiro, simplifique as raízes ao máximo:

     $\mathsf{\big(\sqrt{6}+\sqrt{32}\big)\cdot \big(\sqrt{2}-\sqrt{24}\big)}\\\\ =\mathsf{\big(\sqrt{6}+\sqrt{2^5}\big)\cdot \big(\sqrt{2}-\sqrt{2^3\cdot 3}\big)}\\\\ =\mathsf{\big(\sqrt{6}+\sqrt{2^4\cdot 2}\big)\cdot \big(\sqrt{2}-\sqrt{2^2\cdot 2\cdot 3}\big)}\\\\ =\mathsf{\big(\sqrt{6}+\sqrt{2^{2\,\cdot\,2}\cdot 2}\big)\cdot \big(\sqrt{2}-\sqrt{2^2\cdot 6}\big)}\\\\ =\mathsf{\big(\sqrt{6}+\sqrt{(2^2)^2\cdot 2}\big)\cdot \big(\sqrt{2}-\sqrt{2^2\cdot 6}\big)}\\\\ =\mathsf{\big(\sqrt{6}+\sqrt{(2^2)^2}\cdot \sqrt{2}\big)\cdot \big(\sqrt{2}-\sqrt{2^2}\cdot \sqrt{6}\big)}\\\\ =\mathsf{\big(\sqrt{6}+2^2\cdot \sqrt{2}\big)\cdot \big(\sqrt{2}-2\cdot \sqrt{6}\big)}\\\\ =\mathsf{\big(\sqrt{6}+4\sqrt{2}\big)\cdot \big(\sqrt{2}-2\sqrt{6}\big)}$


Aplicando a distributiva para eliminar os parênteses,
 
     $=\mathsf{\big(\sqrt{6}+4\sqrt{2}\big)\cdot \sqrt{2}-\big(\sqrt{6}+4\sqrt{2}\big)\cdot 2\sqrt{6}}\\\\ =\mathsf{\sqrt{6}\cdot \sqrt{2}+4\sqrt{2}\cdot \sqrt{2}-\sqrt{6}\cdot 2\sqrt{6}-4\sqrt{2}\cdot 2\sqrt{6}}\\\\ =\mathsf{\sqrt{6\cdot 2}+4\sqrt{2\cdot 2}-2\sqrt{6\cdot 6}-8\sqrt{2\cdot 6}}\\\\ =\mathsf{\sqrt{12}+4\sqrt{2^2}-2\sqrt{6^2}-8\sqrt{12}}\\\\ =\mathsf{\sqrt{12}+4\cdot 2-2\cdot 6-8\sqrt{12}}\\\\ =\mathsf{8-12+\sqrt{12}-8\sqrt{12}}\\\\ =\mathsf{-4-7\sqrt{12}}\\\\ =\mathsf{-4-7\sqrt{2^2\cdot 3}}\\\\ =\mathsf{-4-7\sqrt{2^2}\cdot \sqrt{3}}\\\\ =\mathsf{-4-7\cdot 2\cdot \sqrt{3}}$

     $=\mathsf{-4-14\sqrt{3}}$    <———    esta é a resposta.


Bons estudos! :-)