1) 3^12*(3^4)^5/(3^15)^2 2) 6^12/(2^5*3^7)^2 3) - вопрос №68085391312345315226122537230125122374254512535 от mludap 06.10.2020 01:08

Question

Алгебра: 1) 3^12*(3^4)^5/(3^15)^2 2) 6^12/(2^5*3^7)^2 3) 0.125^12*2^37 4)25^4*5/125^3 5)21^8/9^7*7^9 6)(1/3)^17*81^4

mludap · Answer

3^12*(3^4)^5/(3^15)^2= 3^12*3^20/3^30 = 3^32/3^30 = 3^(32-30)=3^2=9

6^12/(2^5*3^7)^2 = (3*2)^12/(2^10*3^14) = 3^(-2)*2^2=(2/3)^2=4/9

0.125^12*2^37 = (2^(-3))^12*2^37=2^(-36)*2^37=2

25^4*5/125^3 = 5^9/5^(3*3)=1

21^8/9^7*7^9 = 3^8*7^8/(3^14*7^9)=1/(3^6*7)=1/5103

(1/3)^17*81^4 = 3^(-17)*(3^16)=1/3