50б помогите пожалуйста с заданием: сократить дробь

Решите задачу:

$\displaystyle\tt1) \ \ \bigg(1\frac{11}{25}\bigg)^6 \cdot\frac{5^{15}}{6^{10}}= \bigg(\frac{36}{25}\bigg)^6 \cdot\frac{5^{15}}{6^{10}}= \frac{(6^2)^6}{(5^2)^6}\cdot\frac{5^{15}}{6^{10}}=\frac{6^{12}}{5^{12}}\cdot\frac{5^{15}}{6^{10}}=6^2\cdot5^3=\\\\\\{} \ \ \ =36\cdot125=4500$

$\displaystyle\tt2) \ \ \bigg(12\frac{1}{4}\bigg)^4 \cdot\frac{2^{9}}{7^{5}}=\bigg(\frac{49}{4}\bigg)^4 \cdot\frac{2^{9}}{7^{5}}=\frac{(7^2)^4}{(2^2)^4}\cdot\frac{2^{9}}{7^{5}}=\frac{7^8}{2^8}\cdot\frac{2^{9}}{7^{5}}=7^3\cdot2=\\\\\\{} \ \ \ =343\cdot2=686$

$\displaystyle\tt3) \ \ \frac{4^{2n-2}\cdot(\frac{1}{6})^{-1-n}}{96^{n-1}} =\frac{4^{2n-2}\cdot6^{-(-1-n)}}{(6\cdot4^2)^{n-1}} =\frac{4^{2n-2}\cdot6^{n+1}}{6^{n-1}\cdot4^{2n-2}} =6^2=36$

$\displaystyle\tt4) \ \ \frac{6^{132}}{4\cdot3^{130}\cdot2^{129}}=\frac{(2\cdot3)^{132}}{2^2\cdot3^{130}\cdot2^{129}}=\frac{2^{132}\cdot3^{132}}{3^{130}\cdot2^{131}}=2\cdot3^2=2\cdot9=18$