tex]...

July 2022 1 8 Report

Упростите выражение:
[tex] \frac{x-0,(3)}{ \sqrt[3]{ x^{2} }+ \sqrt[3]{0,(3)x}+ \sqrt[3]{0,(3)^{2}} } [/tex]

Варианты ответов:
A)[tex] \sqrt[3]{x} - \sqrt[3]{3} [/tex] B)[tex] \sqrt[3]{x} + \sqrt[3]{3} [/tex] C)[tex] \sqrt[3]{x} - \sqrt[3]{3^{-1} } [/tex] D)[tex] \sqrt[3]{x} + \sqrt[3]{3^{-1} } [/tex]

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NNNLLL54

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$\frac{x-0,(3)}{ \sqrt[3]{x^2} +\sqrt[3]{0,(3)x}+\sqrt[3]{0,(3)^2}} =\Big [\; 0,(3)=\frac{3}{9}=\frac{1}{3}\; \Big ]= \frac{x-\frac{1}{3}}{\sqrt[3]{x^2}+\sqrt[3]{\frac{x}{3}}+\sqrt[3]{(\frac{1}{3})^2}} =\\\\\\= \frac{3x-1}{3\cdot (\sqrt[3]{x^2}+\frac{\sqrt[3]{x}}{\sqrt[3]3}+\frac{1}{\sqrt[3]{9}})} = \frac{(\sqrt[3]{3x})^3-1^3}{3\cdot \frac{\sqrt[3]{9x^2}+\sqrt[3]{3x}+1}{\sqrt[3]9} } = \frac{(\sqrt[3]{3x}-1)(\sqrt[3]{9x^2}+\sqrt[3]{3x}+1)}{\sqrt[3]3\cdot (\sqrt[3]{9x^2}+\sqrt[3]{3x}+1)} =$

$= \frac{\sqrt[3]{3x}-1}{\sqrt[3]3} = \frac{\sqrt[3]{3x}}{\sqrt[3]3} - \frac{1}{\sqrt[3]3} = \frac{\sqrt[3]3\cdot \sqrt[3]{x}}{\sqrt[3]3} -(\sqrt[3]3)^{-1}=\sqrt[3]{x}-\sqrt[3]{3^{-1}}\\\\Otvet:\; \; C)\; .$

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