mik ngĩ chắc là = nhau 

Ta có: \(3^{2^3}\)\(=3^8\)

          \(3^{3^2}=3^9\)

Vì 3⁸<3⁹

Nên \(3^{2^3}< 3^{3^2}\)

bằng nhau bạn à

Ta có A = 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰

=> 2A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹

Khi đó 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹) - (1 + 2 + 2² + 2³ + ... + 2¹⁰⁰)

             => A  = 2¹⁰¹ - 1 

Vì 2¹⁰¹ - 1 < 2¹⁰¹

=> A < B

Vậy A < B

A = 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰

=> 2A = 2( 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰ )

           = 2 + 2² + 2³ + ... + 2¹⁰¹

=> A = 2A - A

         = 2 + 2² + 2³ + ... + 2¹⁰¹ - ( 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰ )

         = 2 + 2² + 2³ + ... + 2¹⁰¹ - 1 - 2 - 2² - 2³ - ... - 2¹⁰⁰ 

         = 2¹⁰¹ - 1 < 2¹⁰¹

=> A < B

Ta có :

    \(B=4+2^2+2^3+2^4+...+2^{2016}\)

\(\Rightarrow\) \(B-4=2^2+2^3+2^4+...+2^{2016}\)

\(\Rightarrow\) \(2\left(B-4\right)=2^3+2^4+2^5+...+2^{2017}\)

\(\Rightarrow\) \(2\left(B-4\right)-\left(B-4\right)=B-4=2^{2017}-2^2\)

\(\Rightarrow\) \(B=2^{2017}-2^2+4=2^{2017}\)

\(\Rightarrow\) \(A=B=2^{2017}\)

Vậy \(A=B\)

\(A=\dfrac{2^{2008}-3}{2^{2007}-1};B=\dfrac{2^{2007}-3}{2^{2006}-1}\)

\(\dfrac{1}{2}A=\dfrac{2^{2008}-3}{2^{2008}-2}=1-\dfrac{1}{2^{2008}-2};\dfrac{1}{2}B=\dfrac{2^{2007}-3}{2^{2007}-2}=1-\dfrac{1}{2^{2007}-2}\)

2^2008-2>2^2007-2

=>1/2^2008-2<1/2^2007-2

=>A>B

a) Ta có: a = -1/8 = -9/72

b = 2/-9 = -2/9 = -16/72

Ta thấy: -9 > -16 => -9/72 > -16/72

hay a > b

Vậy a > b

b) Ta có: a = 12/15 = 4/5= 16/20

b = -( -3/4 ) = 3/4= 15/20

Ta thấy: 16 > 15 => 16/20 > 15/20

hay a > b

Vậy a > b

c) Ta có: a = -2/3 = -40/60

b = -0,65 = -13/20 = -39/60

Ta thấy: -40 < -39 => -40/60 < -39/60

hay a < b

Vậy a < b

d) Ta có:  a = -21/3 = -7

b = -413% = -4,13

Ta thấy: -7 < -4,13

=> a < b

Vậy a < b

Chuk bn hok tốt! 

\(6^2=36,2^6=64=>6^2< 2^6\)

\(2^{10}=1024>1000\)

\(6^3-4^3=216-64=152,\left(6-4\right)^3=2^3=8=>6^3-4^3>\left(6-4\right)^3\)