Chọn B

b

a: Ta có: \(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=3\cdot\left(2+2^3+...+2^{99}\right)⋮3\)

b: Ta có: \(B=4+4^2+4^3+...+4^{2022}\)

\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)\)

\(=5\cdot\left(4+4^3+...+4^{2021}\right)⋮5\)

a)\(A=2+2^2+2^3+2^4+...+2^{100}=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)=3.2+3.2^3+...+3.2^{99}=3\left(2+2^3+...+2^{99}\right)⋮3\)b) \(B=4+4^2+4^3+...+4^{2022}=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)=5.4+5.4^3+...+5.4^{2021}=5\left(4+4^3+...+4^{2021}\right)⋮5\)

Ta có:

A = 2 + 2² + 2³ + … + 2²⁰¹⁷

2A = 2.( 2 + 2² + 2³ + … + 2²⁰¹⁷)

2A = 2² + 2³ + 2⁴ + … + 2²⁰¹⁸

2A – A = (2² + 2³ + 2⁴ + … + 2²⁰¹⁸) – (2 + 2² + 2³ + … + 2²⁰¹⁷)

 Vậy  A = 2²⁰¹⁸ – 2

Ta có: A = 2 + 22 + 23 + … + 22017

2A = 2.( 2 + 22 + 23 + … + 22017)

2A = 22 + 23 + 24 + … + 22018

2A – A = (22 + 23 + 24 + … + 22018) – (2 + 22 + 23 + … + 22017)

A = 22018 – 2

Vậy A = 22018 – 2

tick cho mink nhé

😊

a: \(A=\dfrac{3}{5}\cdot xy\cdot\dfrac{4}{25}x^2y^4z^2=\dfrac{12}{125}x^3y^5z^2\)

b: Hệ số là 12/125

Bậc là 10

a) 731 – 680 + 19 = 51 + 19

                              = 70

b) 63 x 2 : 7 = 126 : 7

                    = 18

c) 14 x 6 – 29 = 84 – 29

                       = 55

d) 348 + 84 : 6 = 348 + 14

                        = 362

lấy máy tính ra bấm, không rảnh!!!!!!!

lớp 1 đã học cái này rồi à ?

a) \(A=2+2^2+2^3+...+2^{2017}\)

\(2A=2^2+2^3+2^4+...+2^{2018}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)

\(A=2^{2018}-2\)

b) \(C=1+3^2+3^4+...+3^{2018}\)

\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)

\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)

\(8C=3^{2020}-1\)

\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)

\(Toru\)

a: \(3^8:3^4+2^2\cdot2^3\)

=81+32

=123

b: \(3\cdot4^2-2\cdot3^2\)

\(=48-18\)

=30

a, 3⁸: 3⁴+ 2². 2³

= 3^8-4 + 2²⁺³

= 3⁴ + 2⁵

= 81 + 32

= 113

b,  3 . 4²- 2 . 3²

= 3 . 16 - 2 . 9

= 48 - 18

= 30

c,  84 : 4 + 3⁹: 3⁷+ 5⁰

= 84 : 4 + 3² + 1

= 84 : 4 + 9 + 1

= 21 + 9 + 1

= 31

d,  295 - ( 31 - 2² . 5)²

= 295 - ( 31 - 4 . 5 )²

= 295 - ( 31 - 20 )²

= 295 - 11²

= 295 - 121

= 174

e, 500 - {5[409 - (2³ . 3 - 21)² ] + 10³ } : 15

= 500 -  {5[409 - (8 . 3 - 21)² ] + 10³ } : 15

= 500 -  {5[409 - (24 - 21)² ] + 10³ } : 15

= 500 -  {5[409 - 3² ]+ 10³ } : 15

= 500 - {5[409 - 9 ]+ 10³ } : 15

= 500 - {5 . 400 + 1000 } : 15

= 500 - {2000 + 1000} : 15

= 500 - 3000 : 15

= 500 - 200

= 300

g, 5³ . 2 - 100 : 4 + 2³ . 5

= 125 . 2 - 100 : 4 + 8 . 5

= 250 - 25 + 40

= 225 + 40

= 265

h, 205 - [1200 - (4² - 2 . 3)³ ] : 40

= 205 - [ 1200 - ( 16 - 2 . 3 )³ : 40

= 205 - [ 1200 - ( 16 - 6 )³  ] : 40

= 205 - [ 1200 - 10³ ] : 40

= 205 - [ 1200 - 1000 ] : 40

= 205 - 200 : 40

= 205 - 5

= 200

Đây nha bạn!!!

Ta có: \(A=2+2^2+2^3+\cdots+2^{100}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{101}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{101}-2-2^2-\cdots-2^{100}\)

=>\(A=2^{101}-2\)

=>\(A+2=2^{101}\)

=>\(2^{n}=2^{101}\)

=>n=101