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Алгебра

Ianlmeena

5 год назад

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ОТВЕТЫ

Всеволод

Nov 16, 2020

$\sqrt[3]{36}-\sqrt[3]{30}+\sqrt[3]{25};$

$5+\sqrt{15}-\sqrt{10}-\sqrt{6};$

$2) \frac{11}{\sqrt[3]{6}+\sqrt[3]{5}}=\frac{11*((\sqrt[3]{6})^{2}-\sqrt[3]{6}*\sqrt[3]{5}+(\sqrt[3]{5})^{2})}{(\sqrt[3]{6}+\sqrt[3]{5})*((\sqrt[3]{6})^{2}-\sqrt[3]{6}*\sqrt[3]{5}+(\sqrt[3]{5})^{2})}=\frac{11*(\sqrt[3]{6^{2}}-\sqrt[3]{6*5}+\sqrt[3]{5^{2}})}{(\sqrt[3]{6})^{3}+(\sqrt[3]{5})^{3}}=$

$=\frac{11*(\sqrt[3]{36}-\sqrt[3]{30}+\sqrt[3]{25})}{6+5}=\frac{11*(\sqrt[3]{36}-\sqrt[3]{30}+\sqrt[3]{25})}{11}=\sqrt[3]{36}-\sqrt[3]{30}+\sqrt[3]{25};$

$4) \frac{6}{\sqrt{10}-\sqrt{6}+5-\sqrt{15}}=\frac{6}{\sqrt{2}(\sqrt{5}-\sqrt{3})+\sqrt{5}(\sqrt{5}-\sqrt{3})}=\frac{6}{(\sqrt{2}+\sqrt{5})(\sqrt{5}-\sqrt{3})}=$

$=\frac{6*(\sqrt{2}-\sqrt{5})*(\sqrt{5}+\sqrt{3})}{((\sqrt{2}+\sqrt{5})*(\sqrt{2}-\sqrt{5}))((\sqrt{5}-\sqrt{3})*(\sqrt{5}+\sqrt{3}))}=\frac{6*(\sqrt{10}+\sqrt{6}-5-\sqrt{15})}{((\sqrt{2})^{2}-(\sqrt{5})^{2})((\sqrt{5})^{2}-(\sqrt{3})^{2})}=$

$=\frac{6*(\sqrt{10}+\sqrt{6}-5-\sqrt{15})}{(2-5)(5-3)}=\frac{6*(\sqrt{10}+\sqrt{6}-5-\sqrt{15})}{-3*2}=\frac{6*(\sqrt{10}+\sqrt{6}-5-\sqrt{15})}{-6}=$

$=-(\sqrt{10}+\sqrt{6}-5-\sqrt{15})=5+\sqrt{15}-\sqrt{10}-\sqrt{6};$

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