a) (1+5+5²+5³+...5²⁹⁾chia hết cho 6

Đặt (1+5+5²+5³+...5²⁹) = A

\(A=\left(1+5\right)+\left(5^2+5^3\right)+\left(5^4+5^5\right)....+\left(5^{28}+5^{29}\right)\)

\(A=\left(1+5\right)+5^2\left(5+1\right)+5^4\left(5+1\right)+...+5^{28}\left(5+1\right)\)

\(A=6+5^2.6+5^4.6+...+5^{28}.6\)

Vậy A chia hết cho 6

b) (1+3+3^2+3^3+...+3^29) chia hết cho 13

Đặt B= (1+3+3^2+3^3+...+3^29)

\(B=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{27}+3^{28}+3^{29}\right)\)

\(B=13+3^3\left(1+3+3^2\right)+....+3^{27}\left(1+3+3^2\right)\)

\(B=13+3^3.13+....+3^{27}.13\)

Vậy B chia hết 13

Câu c,d tương tự.Chúc bạn học tốt

A = 2 + 2² + ... + 2¹²⁰

Chứng minh chia hết cho 3

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 ) + 2³( 1 + 2 ) + ... + 2¹¹⁹( 1 + 2 )

= 2.3 + 2³.3 + ... + 2¹¹⁹.3

= 3( 2 + 2³ + ... + 2¹¹⁹ ) chia hết cho 3 ( đpcm )

Chứng minh chia hết cho 7

A = ( 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ ) + ... + ( 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 + 2² ) + 2⁴( 1 + 2 + 2² ) + ... + 2¹¹⁸( 1 + 2 + 2² )

= 2.7 + 2⁴.7 + ... + 2¹¹⁸.7

= 7( 2 + 2⁴ + ... + 2¹¹⁸ ) chia hết cho 7 ( đpcm )

Chứng minh chia hết cho 15

A = ( 2 + 2² + 2³ + 2⁴ ) + ( 2⁵ + 2⁶ + 2⁷ + 2⁸ ) + ... + ( 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 + 2² + 2³ ) + 2⁵( 1 + 2 + 2² + 2³ ) + ... + 2¹¹⁷( 1 + 2 + 2² + 2³ )

= 2.15 + 2⁵.15 + ... + 2¹¹⁷.15

= 15( 2 + 2⁵ + ... + 2¹¹⁷ ) chia hết cho 15 ( đpcm )

1) Ta có: \(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\)

\(A=3\left(2+2^3+...+2^{119}\right)\) chia hết cho 3

2) Ta có: \(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)

\(A=7\left(2+2^4+...+2^{118}\right)\) chia hết cho 7

3) Ta có: \(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(A=2\left(1+2+2^2+2^3\right)+...+2^{117}\left(1+2+2^2+2^3\right)\)

\(A=15\left(2+2^5+...+2^{117}\right)\) chia hết cho 15

Mình chỉ làm được ý 3 thôi: 

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

Chứng minh chia hết cho 7

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

A = (2¹ + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ................ + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ................. + 2¹¹⁸.(1 + 2 + 4)

A = 2.7 + 2⁴ . 7 + ................ + 2¹¹⁸.7

A = 7.(2 + 2⁴ + ........... + 2¹¹⁸)

Chứng minh chia hết cho 31

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰ 

A = (2¹ + 2² + 2³ + 2⁴ + 2⁵) + (2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰) + ................ + (2¹¹⁶ + 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4 + 8 + 16) + 2⁶.(1 + 2 +4 + 8 + 16) + ............. + 2¹¹⁶.(1 + 2 + 4 + 8 + 16)

A = 2.31 + 2⁶.31 + ....... + 2¹¹⁶ . 31

A = 31.(2 + 2⁶ + ........... + 2¹¹⁶)

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

Chứng minh chia hết cho 7

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

A = (2¹ + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ................ + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ................. + 2¹¹⁸.(1 + 2 + 4)

A = 2.7 + 2⁴ . 7 + ................ + 2¹¹⁸.7

A = 7.(2 + 2⁴ + ........... + 2¹¹⁸)

Chứng minh chia hết cho 31

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰ 

A = (2¹ + 2² + 2³ + 2⁴ + 2⁵) + (2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰) + ................ + (2¹¹⁶ + 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4 + 8 + 16) + 2⁶.(1 + 2 +4 + 8 + 16) + ............. + 2¹¹⁶.(1 + 2 + 4 + 8 + 16)

A = 2.31 + 2⁶.31 + ....... + 2¹¹⁶ . 31

A = 31.(2 + 2⁶ + ........... + 2¹¹⁶)

Bài 1:

a: 76-6(x-1)=10

\(\Leftrightarrow x-1=11\)

hay x=12

c: \(5x+15⋮x+2\)

\(\Leftrightarrow x+2=5\)

hay x=3

Bài 1:

a) 76 - 6 (x - 1) = 10

           6 (x - 1) = 76 - 10

           6 (x - 1) = 66

               x - 1 = 66 : 6

               x - 1 = 11

               x      = 11 + 1

               x = 12

b) 3 . 4³ - 7 - 185

= 3 . 64 - 7 - 185

= 192 - 7 - 185

= 185 - 185

= 0

x-15=555

nhóm 4 số đầu tiên vs nhau rồi cứ thế mà đtặ chung

1 + 2 + 2² + ... + 2¹²⁰

= ( 1 + 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ + 2⁷ ) + ... + ( 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 15 + 2⁴(1+2+2²+2³) + ... + 2¹¹⁷(1+2+2²+2³)

= 15.(2⁴+2⁵+...+2¹¹⁷) chia hết cho 15

=> đpcm

ko biết

Bài 1 :

\(A=2+2^2+2^3+2^4+...+2^{118}+2^{119}+2^{120}\)

\(\Rightarrow A=2\left(1+2^{ }+2^2\right)+2^4\left(1+2^{ }+2^2\right)+...+2^{118}\left(1+2^{ }+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7+...+2^{118}.7\)

\(\Rightarrow A=7.\left(2+2^4+...+2^{118}\right)⋮7\)

\(\Rightarrow dpcm\)

Bài 2 :

\(...=23\left(78+22\right)-15=23.100-15=2300-15=2285\)

ko bt bài 1

bài 2 là

=23.(78+22)-15

=23.100-15

=2300-15

=2275

hết

Bài 1 :

A = (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ... + 2¹¹⁸.(1 + 2 + 4)

A = 7.(2 + 2⁴ + ...  + 2¹¹⁸)   ⋮  7 (đfcm)

a) \(A=2+2^2+...+2^{120}\)

\(\Rightarrow A=\left(2+2^2\right)+...+\left(2^{119}+2^{120}\right)\)

\(\Rightarrow A=\left(2+2^2\right)+...+2^{118}.\left(2+2^2\right)\)

\(\Rightarrow A=6+...+2^{118}.6\)

\(\Rightarrow A=6.\left(1+...+2^{118}\right)⋮3\Rightarrow A⋮3\left(đpcm\right)\)

b) \(A=2+2^2+...+2^{120}\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+2^{117}.\left(2+2^2+2^3\right)\)

\(\Rightarrow A=14+...+2^{117}.14\)

\(\Rightarrow A=14.\left(1+...+2^{117}\right)⋮7\Rightarrow A⋮7\left(đpcm\right)\)