Question Number 204841 by Simurdiera last updated on 28/Feb/24
$${Racionalizar}\:{el}\:{denominador} \\ $$$$\frac{\mathrm{1}\:−\:{x}}{\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\sqrt{\sqrt{{x}}}} \\ $$
Commented by Frix last updated on 28/Feb/24
$$\frac{\mathrm{1}−{x}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{4}}} }=−\left(\frac{{x}^{\frac{\mathrm{11}}{\mathrm{12}}} +{x}^{\frac{\mathrm{5}}{\mathrm{6}}} +{x}^{\frac{\mathrm{3}}{\mathrm{4}}} }{{x}}+\underset{{k}=\mathrm{0}} {\overset{\mathrm{8}} {\sum}}{x}^{\frac{{k}}{\mathrm{12}}} \right) \\ $$
Answered by A5T last updated on 28/Feb/24
$$\frac{\left({x}−\mathrm{1}\right)\left(\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}}\right)={p}}{\:\left(\sqrt[{\mathrm{4}}]{{x}}−\sqrt[{\mathrm{3}}]{{x}}\right)\left(\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}}\right)}=\frac{{p}\left(\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)}{\:\left(\sqrt{{x}}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)\left(\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{p}\left(\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)}{{x}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} }={x}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\right)}=\frac{{p}\left(\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{{x}}+\mathrm{1}\right)}{{x}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{{x}}+\mathrm{1}\right)} \\ $$$$=\frac{\left({x}−\mathrm{1}\right)\left(\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}}\right)\left(\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{{x}}+\mathrm{1}\right)}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{−\left(\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}}\right)\left(\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{{x}}+\mathrm{1}\right)}{{x}} \\ $$