Ta có :

A = 2 + 2² + ... + 2²⁰¹⁰

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2²⁰⁰⁹ + 2²⁰¹⁰ )

A = 2 . ( 1 + 2 ) + 2³ . ( 1 + 2 ) + ... + 2²⁰⁰⁹ . ( 1 + 2 )

A = 2 . 3 + 2³ . 3 + ... + 2²⁰⁰⁹ . 3

A = 3 . ( 2 + 2³ + ... + 2²⁰⁰⁹ ) \(⋮\)3

A = 2 + 2² + ... + 2²⁰¹⁰

A = ( 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ ) + ... + ( 2²⁰⁰⁸ + 2²⁰⁰⁹ + 2²⁰¹⁰ )

A = 2 . ( 1 + 2 + 2² ) + 2⁴ . ( 1 + 2 + 2² ) + ... + 2²⁰⁰⁸ . ( 1 + 2 + 2² )

A = 2 . 7 + 2⁴ . 7 + ... + 2²⁰⁰⁸ . 7

A = 7 . ( 2+ 2⁴ + ... + 2²⁰⁰⁸ ) \(⋮\)7

B = 3 + 3² + ... + 3²⁰¹⁰

B = ( 3 + 3² ) + ... + ( 3²⁰⁰⁹ + 3²⁰¹⁰ ) 

Làm tương tự chứng minh được B \(⋮\)4

B = 3 + 3² + ... + 3²⁰¹⁰

B = ( 3 + 3² + 3³ ) + ... + ( 3²⁰⁰⁸ + 3²⁰⁰⁹ + 3²⁰¹⁰ )

Làm tương tự chứng minh được B \(⋮\)13

a, \(A=2+2^2+...+2^{2010}\)

\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)

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

\(\Leftrightarrow A=2.3+2^3.3+...+2^{99}.3\)

\(\Leftrightarrow A=3\left(2+2^2+...+2^{99}\right)\)chia hết cho 3 

a) Ta có : A=(2+2³+2⁵)+(2²+2⁴+2⁶)+.....+(2²⁰⁰⁶+2²⁰⁰⁸+2²⁰¹⁰⁾

                  A=2.(1+2²+2⁴)+2².(1+2²+2⁴)+...+2²⁰⁰⁶.(1+2²+2⁴)

                A=2.21            +2².21             +...+2²⁰⁰⁶.21

                A= 21.(2+2²+...+2²⁰⁰⁶)

                A=3.7.(2+2²+....+2²⁰⁰⁶)   chia hết cho cả 3 và 7

b)b₁. Ta có : B=(3+3²)+...+(3²⁰⁰⁹+3²⁰¹⁰)

                    B=3.(1+3)+...+3²⁰⁰⁹.(1+3)

                    B=3.4        +...+3²⁰⁰⁹.4

                    B= 4.(3+...+3²⁰⁰⁹) chia hết cho 4

b₂₎Ta có : B= (3+3²+3³)+...+(3²⁰⁰⁸+3²⁰⁰⁹+3²⁰¹⁰)

                 B=3.(1+3+3²)+...+3²⁰⁰⁸.(1+3+3²)

                 B= 3.13    +.....+3²⁰⁰⁸.13

                 B=13.(3+.....+3²⁰⁰⁸) chia hết cho 13

NHỚ KICK CHO MÌNH NHA

Câu 1: 

$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$

$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$

$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$

-----------------

$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$

$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$

$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$

$=2+7(2^2+2^5+...+2^{2018})$

$\Rightarrow A$ chia $7$ dư $2$.

Câu 2:

$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$

$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$

-------------------

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$

$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)

giúp mình với mình chuẩn bị phải nộp bài rồi T~T 

\(B=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

1.

a) =>y-5=12 : (2x -1)

các Ư (12) thuộc 1;2;3;4;6;12

mà 2x -1 là số lẻ nên 2x - 1=1 hoặc 3

xét 2 trường hợp trên ta đc x =1 hoặc 2

                                             y=17 hoặc 9

b) vì Ư (7) thuộc 1;7

nên nếu x = 1 thì y =7

       nếu x =7 thì y=1

2.                                      CÓ GHI SAI ĐỀ KO ĐÓ 

DÃY 3 + 3^3 +3^3 LÀ SAO???

Không sai mà

THẾ DÃY ĐÓ THẾ NÀO PHIỀN BN VIẾT THÊM MẤY SỐ TIẾP NHÉ

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

`#3107.101107`

\(A=1+3+3^2+3^3+...+3^{101}\)

$A = (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^{99} + 3^{100} + 3^{101}$

$A = (1 + 3 + 3^2) + 3^3 (1 + 3 + 3^2)  + ... + 3^{99}(1 + 3 + 3^2)$

$A = (1 + 3 + 3^2)(1 + 3^3 + ... + 3^{99})$

$A = 13(1 + 3^3 + ... + 3^{99})$

Vì `13(1 + 3^3 + ... + 3^{99}) \vdots 13`

`\Rightarrow A \vdots 13`

Vậy, `A \vdots 13.`

\(A=1+3+3^2+3^3+3^4+3^5+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)

Vì \(13\cdot(1+3^3+3^6...+3^{99}\vdots13\)

nên \(A\vdots13\)

\(\text{#}Toru\)

Ta có: \(M=3^{2012}-3^{2011}+3^{2010}-3^{2009}\)

\(=\left(3^{2012}+3^{2010}\right)-\left(3^{2011}+3^{2009}\right)\)

\(=3^{2010}\cdot\left(3^2+1\right)-3^{2009}\left(3^2+1\right)\)

\(=\left(3^2+1\right)\cdot\left(3^{2010}-3^{2009}\right)\)

\(=10\cdot3^{2009}\cdot\left(3-1\right)⋮10\)(đpcm)