Question

ПОЖАЛУЙСТА ПОМОГИТЕ ДАЮ 30 БАЛЛОВ!!​

Answer 1

1) 3^(25):3^(22)*3^(2)=3^(25-22+2)=3^(5)

2)6^(20)*6^(18):6^(35)=6^(20+18-35)=6^(3)

3)(5/9)^(40):(5/9)^(36)*(5/9)^(2)=(5/9)^(40-36+2)=(5/9)^(6)

4)(-1/2)^(50):(-1/2)^(49)*(-1/2)^(4)=(-1/2)^(50-49+4)=(-1/2)^(5)

5)(1,1)^(17)*(1,1):(1,1)^(16)=(1,1)^(17+1-16)=(1,1)^(2)

6)(-1,3)^(29):(-1,3)^(28)*(-1,3)=(-1,3)^(29-28+1)=(-1,3)^(2)

Answer 2

Ответ:

$\displaystyle 3^{25}:3^{22}\cdot 3^2=3^{25-22+2}=3^5=243\\\\\\6^{20}\cdot 6^{18}:6^{35}=6^{20+18-35}=6^3=216\\\\\\\Big(\frac{5}{9}\Big)^{40}:\Big(\frac{5}{9}\Big)^{36}\cdot \Big(\frac{5}{9}\Big)^{2}=\Big(\frac{5}{9}\Big)^{40-36+2}=\Big(\frac{5}{9}\Big)^{6}=\frac{5^6}{9^6}=\frac{15625}{531441}$

$\displaystyle \Big(-\frac{1}{2}\Big)^{50}:\Big(-\frac{1}{2}\Big)^{49}\cdot \Big(-\frac{1}{2}\Big)^{4}=\Big(-\frac{1}{2}\Big)^{50-49+4}=\Big(-\frac{1}{2}\Big)^{5}=-\frac{1}{2^5}=-\frac{1}{32}$

$(1,1)^{17}\cdot (1,1):(1,1)^{16}=(1,1)^{17+1-16}=(1,1)^{2}=1,21\\\\\\(-1,3)^{29}:(-1,3)^{28}\cdot (-1,3)=(-1,3)^{29-28+1}=(-1,3)^{2}=1,69$