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Question Number 204841 by Simurdiera last updated on 28/Feb/24

$$Racionalizareldenominador$$$$\frac{1-x}{\sqrt[3]{x}-\sqrt{\sqrt{x}}}$$

Commented by Frix last updated on 28/Feb/24

$$\frac{1-x}{x^{\frac{1}{3}}-x^{\frac{1}{4}}}=-(\frac{x^{\frac{11}{12}}+x^{\frac{5}{6}}+x^{\frac{3}{4}}}{x}+\underset{k=0}{\sum ^8}{x}^{\frac{k}{12}})$$

Answered by A5T last updated on 28/Feb/24

$$\frac{(x-1)(\sqrt[4]{x}+\sqrt[3]{x})=p}{(\sqrt[4]{x}-\sqrt[3]{x})(\sqrt[4]{x}+\sqrt[3]{x})}=\frac{p(\sqrt{x}+\sqrt[3]{x^2})}{(\sqrt{x}-\sqrt[3]{x^2})(\sqrt{x}+\sqrt[3]{x^2})}$$$$=\frac{p(\sqrt{x}+\sqrt[3]{x^2})}{x-\sqrt[3]{x^4}=x(1-\sqrt[3]{x})}=\frac{p(\sqrt{x}+\sqrt[3]{x^2})(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x(1-\sqrt[3]{x})(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}$$$$=\frac{(x-1)(\sqrt[4]{x}+\sqrt[3]{x})(\sqrt{x}+\sqrt[3]{x^2})(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x(1-x)}$$$$=\frac{-(\sqrt[4]{x}+\sqrt[3]{x})(\sqrt{x}+\sqrt[3]{x^2})(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x}$$

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