Question

Racionalizando-se o denominador da fração, obtemos:

Answer 1

Lembre-se que: $\sqrt[b]{a^c}=a^{\frac{c}{b}}$

Assim:

$\frac{a}{ \sqrt[n]{a^{n-2}} }.\frac{\sqrt[n]{a^{n-2}} }{\sqrt[n]{a^{n-2}} } \\\\\\\frac{a\sqrt[n]{a^{n-2}} }{\sqrt[n]{a^{n-2}} ^2} \\\\\\\frac{a.a^{\frac{n-2}{n}}}{\left(a^{\frac{n-2}{n}}\right)^2}\\\\\\\frac{a.a^{\frac{n-2}{n}}}{a^{\frac{2(n-2)}{n}}}\\\\\\\frac{a^{\frac{n-2}{n}+1}}{a^{\frac{2(n-2)}{n}}}\\\\\\\frac{a^{\frac{n-2+n}{n}}}{a^{\frac{2(n-2)}{n}}}\\\\\\\frac{a^{\frac{2n-2}{n}}}{a^{\frac{2(n-2)}{n}}}\\\\\\a^{\frac{2n-2}{n}}.a^{-\frac{2n-4)}{n}}\\\\\\a^{\frac{2n-2}{n}-\frac{2n-4)}{n}}$

$a^{\frac{2}{n}}\\\\\\\sqrt[n]{a^2}$