(1÷15)^(1+2√5)×5^(2+2√5)÷3^(2−2√5)

Решите задачу:

$\displaystyle \frac{\displaystyle \left(\frac{1}{15}\right)^{1+2\sqrt 5}\times5^{2+2\sqrt 5}}{3^{2-2\sqrt 5}}=\frac{\displaystyle \left(\frac{1}{3\cdot5}\right)^{1+2\sqrt 5}\times5^{2+2\sqrt 5}}{3^{2-2\sqrt 5 }}=\\ \\ \frac{ \displaystyle \left(\frac{1}{3\cdot 5}\right)\left( \frac{1}{3\cdot 5} \right)^{2\sqrt 5}\cdot 5^2\cdot 5^{2\sqrt 5} } {\displaystyle \frac{3^2}{3^{2\sqrt 5}}}=\frac{5^2\cdot 5^{2\sqrt 5}\cdot 3^{2\sqrt 5}}{3\cdot 5\cdot 3^{2\sqrt 5}\cdot 5^{2\sqrt 5}\cdot 3^2}=\frac{5}{27}$