ta có: 3²⁰¹⁰ + 5²⁰¹⁰ = (3³)⁶⁷⁰ + (5²)¹⁰⁰⁵ = 27⁶⁷⁰ + 25¹⁰⁰⁵ = (26 + 1)⁶⁷⁰ + (26 - 1)¹⁰⁰⁵ = 26A + 1⁶⁷⁰ - 1¹⁰⁰⁵ = 26A chia hết cho 13

=> 3²⁰¹⁰ + 5²⁰¹⁰ chia hết cho 13

t i c k nha!!!  6756845645765576599435256344465757686878976

=1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 :}

xl mink gần ra oy 

Minh lam cau A) thoi duoc hong

A=2^1+2^2+2^3+2^4+...+2^2010 

=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)

=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)

=2.3+2^3.3+...+2^2010.3

=(2+2^3+2^2010).3

=> A chia het cho 3

​​​​ 

Mà câu c bạn đánh chia hết thành chết hết rồi kìa

em chịu!!!!!!!!!!!

\(3^3=27\equiv1\left(mod13\right)\Rightarrow\left(3^3\right)^{670}\equiv1^{670}\equiv1\left(mod13\right)\) 

\(\equiv5^2=25\equiv-1\left(mod13\right)\Rightarrow\left(5^2\right)^{1005}\equiv\left(-1\right)^{1005}\left(mod13\right)\) 

\(\Rightarrow3^{2010}+5^{2010}\equiv\left(-1\right)+1\equiv0\left(mod13\right)\Rightarrowđpcm\)

32010=(33)670≡1670(mod13)" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">32010=(33)670≡1670(mod13)
52010=(52)1005≡(−1)1005(mod13)" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-table; float:none; font-family:helvea,arial,sans-serif; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">52010=(52)1005≡(−1)1005(mod13)
32010+52010" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">32010+52010 chia hết cho 13

Ta có \(3^{2010}=\left(3^3\right)^{670}=1^{670}\left(mod13\right)\)

Mà \(5^{2010}=\left(5^2\right)^{1005}=\left(-1\right)^{1005}\left(mod13\right)\)

\(\Rightarrow3^{2010}+5^{2010}⋮13\)

A=2010+2010²+2010³+.....+2010²⁰¹⁰

A=2010(1+2010)+2010³(1+2010)+........+2010⁹(1+2010)

A=2010.2011+2010³.2010+....+2010⁹.2011

A=2011(2010+....+2010⁹) chia hết cho 2011

=> A chia hết cho 2011(đpcm)

A = 2010 + 2010² + 2010³ + ... + 2010²⁰¹⁰

A = 2010 . ( 1 + 2010 ) + 2010³ . (1 + 2010 ) + ... + 2010⁹ . ( 1 + 2010 )

A = 2010 . 2011 + 2010³ . 2011 + ... + 2010⁹ . 2011

A = 2011 . ( 2010 + 2010³ + ... + 2010⁹ )

Mà 2011 . ( 2010 + 2010³ + ... + 2010⁹ ) \(\in\)2011

=> A \(\in\)2011

๖²⁴ʱ𝒄𝒉𝒖́𝒄 𝒆𝒎 𝒉𝒐̣𝒄 𝒕𝒐̂́𝒕✟ᴾᴿᴼシ

Lộn ko phải \(\in\)mà là \(⋮\) nha

Bài 1:
$A=2^1+2^2+2^3+2^4$

$2A=2^2+2^3+2^4+2^5$

$\Rightarrow 2A-A=2^5-2^1$

$\Rightarrow A=2^5-1=32-1=31$

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

$B=3^1+3^2+3^3+3^4$

$3B=3^2+3^3+3^4+3^5$

$\Rightarrow 3B-B = 3^5-3$

$\Rightarrow 2B = 3^5-3\Rightarrow B = \frac{3^5-3}{2}$

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

$C=5^1+5^2+5^3+5^4$

$5C=5^2+5^3+5^4+5^5$

$\Rightarrow 5C-C=5^5-5$

$\Rightarrow C=\frac{5^5-5}{4}$

Bài 2: Sai đề bạn nhé. Bạn xem lại.