\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)

\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)

\(\Rightarrow A=2^{2011}-1=B\)

\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)

\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)

\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)

A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰

⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹

⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)

= 2²⁰¹¹ - 2⁰

= 2²⁰¹¹ - 1

= B

Vậy A = B

lo chao cau

a, Ta có  10 30 = 10 3 10 = 1000 10

2 100 = 2 10 10 = 1024 10

Vì 1000<1024 nên  1000 10 <  1024 10

Vậy  10 30 <  2 100

b, Ta có:  333 444 = 333 4 111 = 3 . 111 4 111 =  81 . 111 4 111

444 333 = 444 3 111 = 4 . 111 3 111 =  64 . 111 3 111

Vì 81 > 64 và  111 4 > 111 3 nên  81 . 111 4 111 > 64 . 111 3 111

Vậy  333 444 > 444 333

c, Ta có:  21 5 = 3 . 7 15 = 3 15 . 7 15

27 5 . 49 8 = 3 3 5 . 7 2 8 = 3 15 . 7 16

Vì  7 15 < 7 16 nên  3 15 . 7 15 < 3 15 . 7 16

Vậy  21 5 <  27 5 . 49 8

d, Ta có:  3 2 n = 3 2 n = 9 n

2 3 n = 2 3 n = 8 n

Vì 8 < 9 nên  8 n < 9 n n ∈ N *

Vậy  3 2 n >  2 3 n

e, Ta có: 2017.2018 = (2018–1).(2018+1) = 2018.2018+2018.1–1.2018–1.1

=  2018 2 - 1

Vì  2018 2 - 1 < 2018 2 nên 2017.2018< 2018 2

f, Ta có:  100 - 99 2000 = 1 2000 = 1

100 + 99 0 = 199 0 = 1

Vậy  100 - 99 2000 =  100 + 99 0

g, Ta có:  2009 10 + 2009 9 = 2009 9 . 2009 + 1

=  2010 . 2009 9

2010 10 = 2010 . 2010 9

Vì  2009 9 < 2010 9 nên  2010 . 2009 9 <  2010 . 2010 9

Vậy  2009 10 + 2009 9 <  2010 10

a, 3⁶=3.3.3.3.3.3=729

6³=6.6.6=216

729>216 nên 3⁶>6³

b, 2²⁰⁰=2^2.100=(2²)¹⁰⁰=4¹⁰⁰

4¹⁰⁰=4¹⁰⁰ nên 4¹⁰⁰=2²⁰⁰

c, 333⁴⁴⁴=333^4.111=(333⁴)¹¹¹

444³³³=444^3.111=(444³)¹¹¹

Cả hai số đều cùng có số mũ 111 nên ta so sánh 333⁴ và 444³

333⁴=(3.111)⁴=3⁴.111⁴=81.111⁴

444³=(4.111)³=4³.111³=64.111³

81.111⁴>64.111³ nên 333⁴⁴⁴>444³³³

a, 3⁶ = (3²)³ = 9³ > 6³ vậy 3⁶ > 6³

Các câu khác làm như Lộc 

\(a,3^6=\left(3^2\right)^3=9^3\\ \)

Vì \(9^3>6^3=>3^6>6^3\)

\(b,4^{100}=\left(2^2\right)^{100}=2^{200}\)

Vì \(2^{200}=2^{200}=>4^{100}=2^{200}\)

\(c,333^{444}=111^{444}\cdot3^{444}\\ 444^{333}=111^{333}\cdot4^{333}\\ 3^{444}=\left(3^4\right)^{111}=81^{111}\\4^{333}=\left(4^3\right)^{111}=63^{111} \)

Vì  \(111^{444}>111^{333};81^{111}>63^{111}=>333^{444}>444^{333}\)

\(a.10^{30}=\left(10^3\right)^{10}=1000^{10}\\ 2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì 1000¹⁰ < 1024¹⁰ => 10³⁰ < 2¹⁰⁰

\(b,333^{444}=\left(111\cdot3\right)^{444}=111^{444}\cdot3^{444}=111^{444}\cdot81^{111}\\ 444^{333}=\left(111\cdot4\right)^{333}=111^{333}\cdot4^{333}=111^{333}\cdot64^{111}\)

Vì 111⁴⁴⁴ >111³³³ ; 81¹¹¹ > 64¹¹¹ => 333⁴⁴⁴ > 444³³³

mình cảm ơn

a) >

b) <

c) <

d) <

a)>

b)<

c)<

d)<

a) Cách 1:  2 100 = 2 10 10 = 1024 10 > 1024 9

Cách 2:  1024 9 = 2 10 9 =  2 90  <  2 100

b)  6 . 5 29  >  5 . 5 29  =  5 30

c)  2 98 =  2 2 49 =  4 49  <  9 49

d)  10 30 =  10 3 10 = 1000 10 ;  2 100 =  2 10 10  = 1024 10 nên  10 30  <  2 100

a)  1024 9 = ( 2 10 ) 9 = 2 90 < 2 100

b)  6 . 5 29 > 5 . 5 29 = 5 30

c) 10 30 = ( 10 3 ) 10 = 1000 10 ; 2 100 = ( 2 10 ) 10 = 1024 10   n ê n   10 30 < 2 100 .

a) A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²

2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³

A = 2A - A

= (2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²)

= 2²⁰²³ - 2⁰

= 2²⁰²³ - 1

Vậy A = B

b) A = 2021 . 2023

= (2022 - 1).(2022 + 1)

= 2022.(2022 + 1) - 2022 - 1

= 2022² + 2022 - 2022 - 1

= 2022² - 1 < 2022²

Vậy A < B