mk cx có bài giống cậu đấy

tớ ko giai dai

minh ko giai dai quá  cau phai dang tung cau chứ

a) 5³⁶ = 5¹² (5³)¹² = 125¹²; 11²⁴ = 11^2.12 = (11²)¹² = 121¹²

a: \(4^5\cdot8^7=2^{10}\cdot2^{21}=2^{31}\)

b: \(125^5\cdot25^3=5^{15}\cdot5^6=5^{21}\)

a) \(4^5\)x \(8^7\)

= \(\left(2^2\right)^5\) x \(\left(2^3\right)^7\)

= \(2^{10}\) x \(2^{21}\)

= \(2^{31}\)

b) \(125^5\) x \(25^3\)

= \(\left(5^3\right)^5\) x \(\left(5^2\right)^3\)

= \(5^{15}\) x \(5^6\)

= \(5^{21}\)

a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)

b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)

c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)

\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)

\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)

                       a) \(2^5:8^4=2^5:\left(2^3\right)^4=2^5:2^{12}=2^{-7}=\frac{1}{2^7}\)\(=\left(\frac{1}{2}\right)^7\)

                     b) \(25^6:125^3=\left(5^2\right)^6:\left(5^3\right)^3=5^{12}:5^9=5^3\)

                     c) \(625^2:25^7=\left(5^4\right)^2:\left(5^2\right)^7=5^8:5^{14}=5^{-6}=\frac{1}{5^6}=\left(\frac{1}{5}\right)^6\)

                  d) \(12^3.3^3=\left(12.3\right)^3=36^3\)

                 e) \(32^2:2^4=\left(2^5\right)^2:2^4=2^{10}:2^4=2^6\)

               g) \(64.2:2^5=2^6.2:2^5=2^7:2^5=2^2\)

                Mk làm hộ bn rùi đó nhớ ủng hộ mk nha ^_^

a) (1 phần 2 ) mũ 7

b) 5 mũ 3

c) (1 phần 5 ) mũ 6

d) 36 mũ 3

e) 2 mũ 6

g) 2 mũ 2

Chỉ cần CM 4A:120=>4A +5:125

\(A=5+5^2+5^3+5^4+...+5^{992}\)

\(5A=5^2+5^3+5^4+...+5^{993}\)

\(5A-A=\left(5^2+5^3+5^4+...+5^{993}\right)-\left(5+5^2+5^3+5^4+...+5^{992}\right)\)

\(4A=5^{993}-5\)

\(4A+5=5^{993}\)

\(4A+5=\left(5^3\right)^{331}=125^{331}\)

chịu

Chịu 

\(A=5+5^2+5^3+...+5^{992}\)

\(\Rightarrow5A=5^2+5^3+5^4+...+5^{993}\)

\(\Rightarrow4A=5A-A=5^2+5^3+5^4+...+5^{993}-5-5^2-5^3-...-5^{992}=5^{993}-5\)

\(\Rightarrow4A+5=5^{993}-5+5=5^{993}=\left(5^3\right)^{331}=125^{331}\) là một lũy thừa của 125