Question

(ALFENAS) - Calculando a.√a^-1.√a^-1.√a^-1 obtém-se:

Answer 1

Duas possíveis formas de resolver. Lembrando que na parte final tem que racionalizar o denominador.

$\sqrt{a^3}=\sqrt{a^2 \cdot a}=\sqrt{a^2} \cdot \sqrt{a}=a\sqrt{a}\\ \\--------------------\\ \\ \sqrt{a^{-1}}\cdot \sqrt{a^{-1}}\cdot \sqrt{a^{-1}} = \sqrt{a^{-1}\cdot a^{-1}\cdot a^{-1}} = \sqrt{a^{-1+(-1)+(-1)}} \\ \\= \sqrt{a^{-1-1-1}} =\sqrt{a^{-3}} = \sqrt{ \frac{1}{a^3}} = \frac{1}{\sqrt{a^3}} = \frac{1}{a\sqrt{a}} \\ \\=\frac{1}{a\sqrt{a}} \cdot\frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a\sqrt{a^2}} = \frac{\sqrt{a}}{a\cdot a} = \frac{\sqrt{a}}{a^2}$


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$\sqrt{a^{-1}}= \sqrt\frac{1}{a}}= \frac{\sqrt{1}}{\sqrt{a}}= \frac{1}{\sqrt{a}} \\ \\---------------\\ \\ \sqrt{a^{-1}}\cdot \sqrt{a^{-1}}\cdot \sqrt{a^{-1}} =\frac{1}{\sqrt{a}}\cdot\frac{1}{\sqrt{a}}\cdot\frac{1}{\sqrt{a}}\\ \\=\frac{1}{\sqrt{a\cdot a}\cdot\sqrt{a}}==\frac{1}{\sqrt{a^2}\cdot\sqrt{a}}=\frac{1}{a\sqrt{a}}\\ \\=\frac{1}{a\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}}{a\sqrt{a\cdot a}}=\frac{\sqrt{a}}{a\sqrt{a^2}}=\frac{\sqrt{a}}{a\cdot a}=\frac{\sqrt{a}}{a^2}$
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Veja se a sua questão seria esta:
$\ \ \ a\cdot \sqrt{a^{-1}\cdot\sqrt{a^{-1}\cdot\sqrt{a^{-1}}}}\\ \\=a\cdot \sqrt{a^{-1}\cdot\sqrt{a^{-1}\cdot a^{- \frac{1}{2} }}}\\ \\=a\cdot \sqrt{a^{-1}\cdot\sqrt{a^{-1- \frac{1}{2} }}}\\ \\=a\cdot \sqrt{a^{-1}\cdot\sqrt{a^{- \frac{3}{2} }}}\\ \\=a\cdot \sqrt{a^{-1}\cdot\big(a^{- \frac{3}{2}}\big)^ \frac{1}{2} }}}$

$=a\cdot \sqrt{a^{-1}\cdot a^{- \frac{3}{4}}}}}\\ \\=a\cdot \sqrt{a^{-1-\frac{3}{4}}}}\\ \\=a\cdot \sqrt{a^{-\frac{7}{4}}}\\ \\=a\cdot \big(a^{-\frac{7}{4}}\big)^{ \frac{1}{2}} \\=a\cdot a^{-\frac{7}{8}}\\ \\= a^{1-\frac{7}{8}}\\ \\= a^{\frac{1}{8}}\\ \\= \sqrt[8]{a}$