Ответ:
Применяем правила \((a^m)^n=a^{mn}\), \(a^m\cdot a^n=a^{m+n}\) и \(a^m:a^n=a^{m-n}\).
- \((a^3)^7\cdot(a^4)^3=a^{21}a^{12}=a^{33}\).
- \((a^2)^3\cdot(a^5)^4=a^6a^{20}=a^{26}\).
- \((a^3)^2\cdot(a^5)^3=a^6a^{15}=a^{21}\).
- \((a^2)^4\cdot(a^3)^5=a^8a^{15}=a^{23}\).
- \((a^4)^3\cdot(a^3)^2=a^{12}a^6=a^{18}\).
- \(\dfrac{4^3\cdot16^2}{2^{12}}=\dfrac{(2^2)^3\cdot(2^4)^2}{2^{12}}=2^{6+8-12}=2^2=4\).
- \(\dfrac{5^9\cdot4^6}{20^6}=\dfrac{5^9\cdot4^6}{(5\cdot4)^6}=5^3=125\).
- \(\dfrac{3^8\cdot7^8}{21^7}=\dfrac{(3\cdot7)^8}{21^7}=21\).
- \(\dfrac{25^3\cdot125^2}{5^{10}}=\dfrac{5^6\cdot5^6}{5^{10}}=5^2=25\).
- \(\dfrac{7^{14}\cdot(7^2)^3}{(7^3)^6\cdot7^2}=7^{14+6-18-2}=7^0=1\).
- \(\dfrac{3^{10}\cdot(3^3)^5}{(3^5)^4\cdot3}=3^{10+15-20-1}=3^4=81\).
