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

Lời giải:

Hiển nhiên $A\vdots 7$ do các số hạng đều chia hết cho 7.

Lại có:

$A=(7+7^3)+(7^5+7^7)+....+(7^{1997}+7^{1999})$

$=7(1+7^2)+7^5(1+7^2)+...+7^{1997}(1+7^2)$
$=(1+7^2)(7+7^5+...+7^{1997})$
$=50(7+7^5+...+7^{1997})\vdots 5$

Vậy $A\vdots 7, A\vdots 5$. Mà $(7,5)=1$

$\Rightarrow A\vdots 35$

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

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

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

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

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

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/ 

\(\overline{aba}=101.a+10b=98a+3a+7b+3b=\)

\(=\left(98a+7b\right)+3\left(a+b\right)\)

\(98a+7b⋮7;\left(a+b\right)⋮7\Rightarrow3\left(a+b\right)⋮7\)

\(\Rightarrow\overline{abc}=\left(98a+7b\right)+3\left(a+b\right)⋮7\)

b/ xem lại đề bài

Ta có :

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

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

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

   = 7 . 50 + 7⁵ . 50 + ... + 7¹⁹⁹⁷ . 50

   = 350 + 7⁴ . 350 + ... + 7¹⁹⁹⁶ . 350

   = 35 . 10 + 7⁴ . 35 . 10 + ... + 7¹⁹⁹⁶ . 35 . 10

   = 35 (10 + 7⁴ . 10 + ... + 7¹⁹⁹⁶ . 10) chia hết cho 35

Vậy A chia hết cho 35 (ĐPCM).

Đáp án của tôi cũng giống như bạn Trần Hùng Minh vậy .

A = 7+7^3+...+7^1999

A = 7.(1+49)+...+7^1997.(1+49)

A = 7.50+...+7^1997.50

A = 350+...+7^1996.7.50

A = 350+...+7^1996.350

A = 350.(1+...+7^1996) = 35.10.(1+...+7^1996) => A chia hết cho 35

A=7+7^3+7^5+..............+7^999

  =[7+7^3]+[7^5+7^7]+..............+[7^997+7^999] 

  =7[1+7^2]+7^5[1+7^2]+..............+7^997[1+7^2]

  =7[1+49]+7^5[1+49]+................7^997[1+49]

  =7*50+7^5*50+...................+7^997*50

  =350+7^4*7*50+.................+7^996*7*50

  =350+7^4*350+................+7^996*350

  =350[1+7^4+................+7^996]

vì 350 chia hết cho 35 nên A chia hết cho 35

\(_{^{ }^{ }^{ }^{ }^{ }^{ }^{ }^{ }^{ }\veebar\circledcircℕ^∗\Phi}\)

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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