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

BÀI BẠN GIỐNG Y CHANG BÀI MIK LUÔN

17/21>17/29>15/29 (tự KL) b)(chép đề) B= (20+21)/(21+22)=41/43<1 (chép đề sai). Xét A= 20/21+21/22=1-1/21+1-1/22=1+1-(1/21+1/22). Ta thấy (1/21+1/22)<1 nên 1-(1/21+1/22)>0

17/21>17/29>15/29 (tự KL)
b)(chép đề)
B= (20+21)/(21+22)=41/43<1 (chép đề sai).
Xét A= 20/21+21/22=1-1/21+1-1/22=1+1-(1/21+1/22). 
Ta thấy (1/21+1/22)<1 
nên 1-(1/21+1/22)>0
Vậy 1+1-(1/21+1/22)>1+0>1
Vậy A>B

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

A Lớn hơn

\(A=\left(\frac{20}{5}+\frac{27}{9}\right)\times\frac{21}{10}=\left(4+3\right)\times\frac{21}{10}=7\times\frac{21}{10}=\frac{147}{10}\)

\(B=\left(\frac{13}{6}-\frac{3}{8}\right)\times\frac{11}{22}\)

\(B=\left(\frac{52}{24}-\frac{9}{24}\right)\times\frac{11}{22}\)

\(B=\frac{43}{24}\times\frac{1}{2}=\frac{43}{48}\)

Dễ thấy \(A=\frac{147}{10}>1\)

Mà \(B=\frac{43}{48}< 1\)

=> tự so sánh

A = \(\dfrac{10^{20}+3}{10^{21^{ }}+3}\)

B = \(\dfrac{10^{21}+4}{10^{22}+4}\) < 1

\(\Rightarrow\) B < \(\dfrac{10^{21}+4+6}{10^{22}+4+6}\) 

\(\Rightarrow\) B < \(\dfrac{10^{21}+10}{10^{22}+10}\)

\(\Rightarrow\) B < \(\dfrac{10\left(10^{20}+1\right)}{10\left(10^{21}+1\right)}\)

\(\Rightarrow\) B < \(\dfrac{10^{20}+1}{10^{21}+1}\) < \(\dfrac{10^{21}+1+2}{10^{22}+1+2}\)

\(\Rightarrow\) B < \(\dfrac{10^{21}+3}{10^{22}+3}\)

\(\Rightarrow\) B < A

Ta có: \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};\ldots;\frac{1}{39}>\frac{1}{40};\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{20}{40}=\frac12\) (1)

Ta có: \(\frac{1}{41}>\frac{1}{80};\frac{1}{42}>\frac{1}{80};\ldots;\frac{1}{79}>\frac{1}{80};\frac{1}{80}=\frac{1}{80}\)

Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+\cdots+\frac{1}{80}=\frac{40}{80}=\frac12\) (2)

Từ (1),(2) suy ra \(B>\frac12+\frac12\)

=>B>1

mà \(1>\frac{39}{40}=A\)

nên B>A

Đề bài yêu cầu gì vậy bạn? Rút gọn ạ?

`@` Đặt `A=2^1+2^2+2^3+...+2^2017`

`=>2A=2(2^1+2^2+2^3+...+2^2017)`

`=>2A=2^2+2^3+...+2^2018`

`=>2A-A=(2^2+2^3+...+2^2018)-(2^1+2^2+...+2^2017)`

`=>A=2^2018-2`