tex]...

August 2022 1 10 Report

Избавьтесь от иррациональной в знаменателе дроби

(photomath не помогает)
[tex] \frac{1}{ \sqrt[3]{3} - 1} [/tex]
[tex] \frac{1}{ \sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4} } [/tex]

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sangers1959

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Объяснение:

$\frac{1}{\sqrt[3]{3}-1 } =\frac{(\sqrt[3]{3})^2+\sqrt[3]{3}*1+1^2 }{(\sqrt[3]{3}-1)*((\sqrt[3]{3})^2+\sqrt[3]{3}*1+1^2 ) } =\frac{\sqrt[3]{9} +\sqrt[3]{3} +1}{(\sqrt[3]{3} )^3-1^3} =\frac{\sqrt[3]{9} +\sqrt[3]{3} +1}{3-1} =\frac{\sqrt[3]{9} +\sqrt[3]{3} +1}{2} .$

$\frac{1}{\sqrt[3]{9} -\sqrt[3]{6} +\sqrt[3]{4} } =\frac{\sqrt[3]{3}+\sqrt[3]{2} }{(\sqrt[3]{3}+\sqrt[3]{2})*(\sqrt[3]{9} -\sqrt[3]{6} +\sqrt[3]{4}) } =\frac{\sqrt[3]{3}+\sqrt[3]{2}}{(\sqrt[3]{3})^3+(\sqrt[3]{2})^3 } =\frac{\sqrt[3]{3}+\sqrt[3]{2}}{3+2}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{5}.$

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