\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)=7.57+7^4.57+...+7^{118}.57=57\left(7+7^4+...+7^{118}\right)⋮57\)

Lời giải:
$A=(7+7^2+7^3)+(7^4+7^5+7^6)+....+(7^{118}+7^{119}+7^{120})$
$=7(1+7+7^2)+7^4(1+7+7^2)+...+7^{118}(1+7+7^2)$

$=7.57+7^4.57+...+7^{118}.57$

$=57(7+7^4+...+7^{118})\vdots 57$ 

Ta có đpcm.

A = 7 + 72 + 73 + ... + 7119 + 7120

A = (71 + 72 + 73) + (74 + 75 + 76) + ... + (7118 + 7119 + 7120)

A = 7(1 + 7 + 72) + 74(1 + 7 + 72) + ... + 7118(1 + 7 + 72)

A = 7.57 + 74.57 + ... + 7118.57

A = 57(7 + 74 + ... + 7118)

Vì 57 ⋮ 57 nên 57(7 + 74 + ... + 7118) ⋮ 57

A = 7 + 72 + 73 + ... + 7119 + 7120

A = (71 + 72 + 73) + (74 + 75 + 76) + ... + (7118 + 7119 + 7120)

A = 7(1 + 7 + 72) + 74(1 + 7 + 72) + ... + 7118(1 + 7 + 72)

A = 7.57 + 74.57 + ... + 7118.57

A = 57(7 + 74 + ... + 7118)

Vì 57 ⋮ 57 nên 57(7 + 74 + ... + 7118) ⋮ 57

\(A=\left(1+7\right)+...+7^{2020}\left(1+7\right)=8\left(1+...+7^{2020}\right)⋮8\)

\(A = (1 + 7) +...+7^2\)\(^0\)\(^2\)\(^0\) \((1 + 7) = 8 (1+...+7^2\)\(^0\)\(^2\)\(^0\)\() \) ⋮\(8\)

đề thiếu bạn ơi

hoặc đề sai

..............

bn viết thiếu đề nhé

A= 71 + 72 + 73 + 74 = (71+74)+(72+73) = 145 + 145 = 290 chia hết cho 5

=> A=........ chia hết cho 5

B=  10⁶-5⁷ = 2⁶. 5⁶ - 5⁷ = 5⁶ ( 2⁶ - 5) =(5⁶ . 59) chia hết cho 59 => B chia hết cho 59

A bn viết thiếu chỗ 7 => 71 đấy

\(A=7+7^2+7^3+...+7^{120}\\ A=\left(7+7^2+7^3\right)+...+\left(7^{118}+7^{119}+7^{120}\right)\\ A=7\times\left(1+7+7^2\right)+...+7^{118}\times\left(1+7+7^2\right)\\ A=7\times57+7^4\times57+...+7^{118}\times57\\ A=57\times\left(7+7^4+...+7^{118}\right)\\ \Rightarrow A⋮57\)

A = 7 + 72 + 73 + ... + 7119 + 7120

A = (71 + 72 + 73) + (74 + 75 + 76) + ... + (7118 + 7119 + 7120)

A = 7(1 + 7 + 72) + 74(1 + 7 + 72) + ... + 7118(1 + 7 + 72)

A = 7.57 + 74.57 + ... + 7118.57

A = 57(7 + 74 + ... + 7118)

Vì 57 ⋮ 57 nên 57(7 + 74 + ... + 7118) ⋮ 57

1.

\(A=7+7^2+7^3+...+7^{78}\)

\(=\left(7+7^2\right)+\left(7^3+7^4\right)+...+\left(7^{77}+7^{78}\right)\)

\(=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{77}\left(1+7\right)\)

\(=7\cdot8+7^3\cdot8+...+7^{77}\cdot8\)

\(=\left(7+7^3+...+7^{77}\right)\cdot8\) chia hết cho 8

Vậy A chia hết cho 8 (đpcm)

\(A=3+3^2+3^3+...+3^{155}\)

\(=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{151}+3^{152}+3^{153}+3^{154}+3^{155}\right)\)

\(=3\left(1+3+3^2+3^3+3^4\right)+...+3^{151}\left(1+3+3^2+3^3+3^4\right)\)

\(=\left(3+...+3^{151}\right)\cdot121\) chia hết cho 121

Vậy A chia hết cho 121 (đpcm)

\(A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8\)

\(A=\left(7+7^3\right)+\left(7^2+7^4\right)+\left(7^5+7^7\right)+\left(7^6+7^8\right)\)

\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)

\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)

\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)

\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)

Ta có: 5 ⋮ 5

⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm) 

A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78

A =  (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)

A = 7.(1 + 72)  + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)

A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)

A = 7.50 + 72.50 + 75.40 + 76.50

A = 50.(7 + 72 + 75 + 76)

Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm

A = 7 + 7² + 7³ + 7⁴ + 7⁵ + 7⁶ + 7⁷ + 7⁸

A =  (7 + 7³) + (7²+ 7⁴) + (7⁵ + 7⁷) + (7⁶ + 7⁸)

A = 7.(1 + 7²)  + 7².(1 + 7²) + 7⁵.(1 + 7²) + 7⁶.(1 + 7²)

A = 7.( 1 + 49) + 7².( 1 + 49) + 7⁵.(1 + 49) + 7⁶. (1 + 49)

A = 7.50 + 7².50 + 7⁵.50 + 7⁶.50

A = 50.(7 + 7² + 7⁵ + 7⁶)

Vì 50 ⋮ 5 nên A = 50.(7 + 7² + 7⁶) ⋮ 5 đpcm

Sai đề ruj A=137256 ko thể chia hết cho 50

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