64^8=(4^3)^8=4^24=16^12>16^10

=>64^8>16^10

\(32^9:64=\left(2^5\right)^9:2^6=2^{45}:2^6=2^{45-6}=2^{39}\)

32⁹:64

= (2⁵)⁹:2⁶

= 2⁴⁵:2⁶

= 2^45-6

= 2³⁹

T/c: \(S = 1+2+2^2+2^3+...+\)\(2^{62}+2^{63}\) (1)

Nhân hai vế với 2 ta đc:

\(2S = 2+2^2+^3+...+\)\(2^{63}+2^{64}\) (2)

Trừ từng vế đẳng thức (2) cho đẳng thức (1), ta có: \(S=2^{64}-1\).

đặt A=1+2+\(2^2\)+........+\(2^{63}\)thì 2A=2+2²+\(2^3\)^+.......+\(2^{64}\)

suy ra 2A-A=\(2^{64}\)-1 hay A=\(2^{64}\)-1

vậy D=\(2^{64}\)-A=\(2^{64}\)-\(2^{64}\)+1=1

8,16,32,64,125,256,512,1024

a) 2³= 8 

, 2⁴=16 

, 2⁵=32 

, 2⁶ =64,

2⁷ =128

2⁸ =256

2⁹ =512

2¹⁰ =1024

b) 3² =9

 3³ =27

 3⁴ =81

 3⁵ =243

c) 4² =16

 4³ =64

4⁴ =256

d) 5² =25

 5³ =125

5⁴ =625

e) 6² = 36

 6³ =216

 6⁴ =1296

hok tốt

Giải:

a) \(4^2.8^5=\left(2^2\right)^2.\left(2^3\right)^5=2^4.2^{15}=2^{19}\)

b) \(27^3.9^4=\left(3^3\right)^3.\left(3^2\right)^4=3^9.3^8=3^{17}\)

c) \(64^3.16^7=\left(4^3\right)^3.\left(4^2\right)^7=4^9.4^{14}=4^{23}\)