Comparação algébrica entre números reais

por math_estudos15 Sex 02 Out 2020, 19:24

Sendo a um número real entre 0 e 1, é CORRETO afirmar que:

A [latex]\sqrt[3]{a^{5}} > \sqrt{a}[/latex]
B [latex]\frac{1}{\sqrt{a^{5}}}<\frac{1}{\sqrt[3]{a^{2}}}[/latex]
C [latex]\sqrt[3]{a}>\sqrt[6]{a^{2}}[/latex]
D [latex]\sqrt[3]{a^{5}}<\sqrt[3]{a^{2}}[/latex]

Spoiler:

por **Elcioschin** Sex 02 Out 2020, 19:32

Se a base está no intervalo]0, 1[, quanto maior o expoente menor a potência

Eis um exemplo da alternativa a) :

∛(a⁵) = a^5/3

√a = a^1/2

5/3 > 1/2 ---> a^5/3 < a^1/2 ---> Falso

Faça os outros

por **Christian M. Martins** Sex 02 Out 2020, 20:52

[latex]\sqrt[3]{a^{5}} = a^{\frac{5}{3}}[/latex]
[latex]\sqrt{a} = a^{\frac{1}{2}}[/latex]

[latex]\frac{5}{3} > \frac{1}{2} \therefore \sqrt[3]{a^{5}} < \sqrt{a}[/latex]

[latex]\frac{1}{\sqrt{a^{5}}} = (\sqrt{a^{5}})^{-1} = a^{\frac{5}{2}^{-1}} = a^{-1(\frac{5}{2})} = a^{-\frac{5}{2}}[/latex]
[latex]\frac{1}{\sqrt[3]{a^{2}}} = (\sqrt[3]{a^{2}})^{-1} = a^{\frac{2}{3}^{-1}} = a^{-1(\frac{2}{3})} = a^{-\frac{2}{3}}[/latex]

[latex]-\frac{5}{2} < - \frac{2}{3} \therefore \frac{1}{\sqrt{a^{5}}} > \frac{1}{\sqrt[3]{a^{2}}}[/latex]

[latex]\sqrt[3]{a} = a^{\frac{1}{3}}[/latex]
[latex]\sqrt[6]{a^{2}} = a^{\frac{2}{6}} = a^{\frac{1}{3}}[/latex]

[latex]\frac{1}{3} = \frac{1}{3} \therefore \sqrt[3]{a} = \sqrt[6]{a^{2}}[/latex]

[latex]\sqrt[3]{a^{5}} = a^{\frac{5}{3}}[/latex]
[latex]\sqrt[3]{a^{2}} = a^{\frac{2}{3}}[/latex]

[latex]\frac{5}{3} > \frac{2}{3} \therefore \sqrt[3]{a^{5}} < \sqrt[3]{a^{2}}[/latex]

Alternativa D.