415. ${{25}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}:{{5}^{{}^{-3}\!\!\diagup\!\!{}_{2}\;}}$ es igual a

a) ${{5}^{4}}\sqrt{5}$

b) ${{5}^{2}}$

c) $125\sqrt[3]{25}$

 

Teniendo en cuenta las propiedades de las potencias de números reales:

 

${{25}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}:{{5}^{{}^{-3}\!\!\diagup\!\!{}_{2}\;}}\quad =\quad \frac{{{25}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}}{{{5}^{{}^{-3}\!\!\diagup\!\!{}_{2}\;}}}\quad =\quad {{25}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}\cdot {{5}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}\quad =\quad {{\left( 25\cdot 5 \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}\quad =$

 

$=\quad {{\left( {{5}^{2}}\cdot 5 \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}\quad =\quad {{\left( {{5}^{2+1}} \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}\quad =\quad {{\left( {{5}^{3}} \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}\quad =\quad {{5}^{3\cdot \frac{3}{2}}}\quad =$

 

$=\quad {{5}^{\frac{9}{2}}}\quad =\quad {{5}^{\frac{8}{2}+\frac{1}{2}}}\quad =\quad {{5}^{\frac{8}{2}}}\cdot {{5}^{\frac{1}{2}}}\quad =\quad {{5}^{4}}\sqrt{5}$