Castillos con potencias

Para realizar estos ejercicios realizamos las operaciones del numerador y del denominador, de la fracción grande, por separado.

Unos cuentos ejemplos resueltos de "castillos con potencias".

$\bullet\dfrac {\left( \dfrac {2} {3}\right) ^{6}\cdot \left( \dfrac {2} {3}\right) ^{7}} {\left( \dfrac {2} {3}\right) ^{3}\cdot\left( \dfrac {2} {3}\right) ^{4}\cdot \left( \dfrac {2} {3}\right) ^{5}}=\dfrac {\left( \dfrac {2} {3}\right) ^{6+7}} {\left( \dfrac {2} {3}\right) ^{3+4+5}}=\dfrac {\left( \dfrac {2} {3}\right) ^{13}} {\left( \dfrac {2} {3}\right) ^{12}}=\left( \dfrac {2} {3}\right) ^{13-12}=\boxed{\dfrac {2} {3}}$

$\bullet\dfrac {\left( \dfrac {1} {2}\right) ^{4}:\left( \dfrac {1} {2}\right) } {\left( \dfrac {1} {2}\right) ^{2}\dfrac {1} {2}}=\dfrac {\left( \dfrac {1} {2}\right) ^{4-1}} {\left( \dfrac {1} {2}\right) ^{2+1}}=\dfrac {\left( \dfrac {1} {2}\right) ^{3}} {\left( \dfrac {1} {2}\right) ^{3}}=\boxed{1}$

$\bullet\dfrac {\left[ \left( \dfrac {2} {5}\right) ^{2}\right] ^{6}\cdot\left[ \left( \dfrac {2} {5}\right) ^{9}\right] ^{4}} {\left( \dfrac {2} {5}\right) ^{40}}=\dfrac {\left( \dfrac {2} {5}\right) ^{2\cdot 6}\cdot \left( \dfrac {2} {5}\right) ^{9\cdot 4}} {\left( \dfrac {2} {5}\right) ^{40}}=\left( \dfrac {2} {5}\right) ^{26+9\cdot 4-40}=\boxed{\left( \dfrac {2} {5}\right) ^{8}}$