A=(7+73)+(75+77)+....+(71997+71999)

A=7.(1+72)+75.(1+72)+....+71997.(1+72)

A=7.50+75.50+79.50+.....+71997.50

=>A chia hết cho 5 (1)

A=(7+73+75+....+71999)=7.(70+72+74+....+71998)

=>A chia hết cho 7 (2)

Mà ƯCLN(5;7)=1=>A chia hết cho 35

Ban kia lam dung roi

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

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

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

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

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

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

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

a: \(=2^2\left(1+2\right)+2^4\left(1+2\right)=3\left(2^2+2^4\right)⋮3\)

b: \(=4^{20}\left(1+4\right)+4^{22}\left(1+4\right)=5\left(4^{20}+4^{22}\right)⋮5\)

c: \(A=\left(1+4+4^2\right)+...+4^{96}\left(1+4+4^2\right)\)

\(=21\left(1+...+4^{96}\right)⋮21\)

d: \(B=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{35}\left(1+7\right)\)

\(=8\left(7+7^3+...+7^{35}\right)⋮8\)

\(B=7\left(1+7+7^2\right)+...+7^{34}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{34}\right)\) chia hếtcho 3 và 19

A=77+105+161

TA THẤY 77 CHIA HẾT CHO 7

               105 CHIA HẾT CHO 7

                161 CHIA HẾT CHO 7

NÊN ĐỂ A CHIA HẾT CHO 7 THÌ X CŨNG PHẢI CHIA HẾT CHO 7 => X THUỘC 7K

NGƯỢC LẠI NẾU ĐỂ A KHÔNG CHIA HẾT CHO 7 => X KHÁC 7K

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

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

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.

 á à thg hếu cx hỏi trên này cơ à XDDD

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:

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

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