7. Вычислите $$\frac{7^4 \cdot 7^8}{7^{10} \cdot 41^{12}} \cdot \frac{41^{12}}{41^2 \cdot 41^9}$$

$$\frac{7^4 \cdot 7^8}{7^{10} \cdot 41^{12}} \cdot \frac{41^{12}}{41^1 \cdot 41^{10}} = \frac{7^{4+8}}{7^{10} \cdot 41^{12}} \cdot \frac{41^{12}}{41^{1+10}} = \frac{7^{12}}{7^{10} \cdot 41^{12}} \cdot \frac{41^{12}}{41^{11}} = 7^{12-10} \cdot 41^{12-12} \cdot 41^{12-11} = 7^2 \cdot 41^0 \cdot 41^1 = 49 \cdot 1 \cdot 41 = 49 \cdot 41 = 2009$$

Ответ: 2009