a: \(B=3+3^2+3^3+...+3^{120}\)

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

b: \(B=3+3^2+3^3+3^4+...+3^{2020}\)

\(=3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)

\(=4\cdot\left(3+...+3^{2019}\right)⋮4\)

\(a,S=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\\ S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ S=\left(3+3^2\right)\left(1+3^2+...+3^{18}\right)=12\left(1+3^2+...+3^{18}\right)⋮12\)

\(b,S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ S=\left(3+3^2+3^3+3^4\right)+....+3^{16}\left(3+3^2+3^3+3^4\right)\\ S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ S=120\left(1+...+3^{16}\right)⋮120\)

\(a,S=3+3^2+3^3+...+3^{20}\)

Ta thấy:\(3+3^2=12⋮12\)

\(\Rightarrow S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ \Rightarrow S=\left(3+3^2\right)\left(1+3^2+...+1^{18}\right)\\ \Rightarrow S=12.\left(1+3^2+...+3^{18}\right)⋮12\\ \left(đpcm\right)\)

\(b,Ta\) \(thấy:\)\(3+3^2+3^3+3^4=120⋮120\)

\(\Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+3^{16}\left(3+3^2+3^3+3^4\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ \Rightarrow S=120\left(1+...+3^{16}\right)⋮120\\ \left(đpcm\right)\)

a) \(\frac{4n+1}{2n-1}=\frac{4n-2+3}{2n-1}=\frac{2.\left(2n-1\right)+3}{2n-1}\)

\(=2+\frac{3}{2n-1}\). Vì \(2\in Z\Rightarrow\frac{3}{2n-1}\in Z\Rightarrow2n-1\inƯ\left(3\right)\)

\(\Rightarrow2n-1\in\left\{-3;-1;1;3\right\}\)

\(\Rightarrow2n\in\left\{-2;0;2;4\right\}\)

\(\Rightarrow n\in\left\{-1;0;1;2\right\}\)

b)\(\frac{2n+5}{n+2}=\frac{2n+4+1}{n+2}=\frac{2.\left(n+2\right)+1}{n+2}\)

\(=\frac{2.\left(n+2\right)}{n+2}+\frac{1}{n+2}=2+\frac{1}{n+2}\). Vì \(2\in Z\Rightarrow n+2\inƯ\left(1\right)\)

\(\Rightarrow n+2\in\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{-3;-1\right\}\)

c) \(\frac{2n-3}{n-2}=\frac{2n-4+1}{n-2}=\frac{2.\left(n-2\right)+1}{n-2}\)

\(=\frac{2.\left(n-2\right)}{n-2}+\frac{1}{n-2}=2+\frac{1}{n-2}\)

Vì \(2\in Z\Rightarrow\frac{1}{n-2}\in Z\Rightarrow n-2\inƯ\left(1\right)\)

\(\Rightarrow n-2\in\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{1;3\right\}\)

Ta có: \(4n+1⋮2n-1\Leftrightarrow4n-2+3⋮2n+1\)\(\Leftrightarrow2\left(2n-1\right)+3⋮2n-1\Leftrightarrow3⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow2n=\left\{-2;0;2;4\right\}\)

Vì \(n\in N\)nên \(n=\left\{0;1;2\right\}\)

Các câu trên bỏ các giá trị của n mà có dấu "-" nha bạn. Mình chưa đọc kỹ đề

B=\(3^1+3^2+3^3+...+3^{300}\)

  =\(\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{299}+3^{300}\right)\) 

  =\(3\left(1+3\right)+3^3\left(1+3\right)+...+3^{299}\left(1+3\right)\)

  =\(3.4+3^3.4+...+3^{299}.4\)

  =\(\left(3+3^3+...+3^{299}\right).4\)

Vì 4\(⋮\)2 mà trong một tích có 1 ts chia hết cho 2 thì tích đó chia hết cho 2 \(\Rightarrow\)B\(⋮\)2

64¹⁰ -32 ¹¹ - 16¹³ = 2⁶⁰ - 2⁵⁵ - 2⁵² = 2⁵² . 2⁸ - 2⁵² . 2³ - 2⁵²

= 2⁵² ( 2⁸ - 2³ - 1) 

= 2⁵² . 247 = 2⁵² . 19 . 13

=> chia hết cho 19           

cảm ơn nhiều ạ

chắc là lớp 8 hay 9 rồi đúng ko ạ ?

64¹⁰ -32 ¹¹ - 16¹³ = 2⁶⁰ - 2⁵⁵ - 2⁵² = 2⁵² . 2⁸ - 2⁵² . 2³ - 2⁵²

= 2⁵² ( 2⁸ - 2³ - 1) 

= 2⁵² . 247 = 2⁵² . 19 . 13

=> chia hết cho 19           

Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

\(a,\)Để \(n+3⋮n\)

Mà \(n⋮n\Rightarrow3⋮n\)

=> n là ước của 3 .

Mà n lại số tự nhiên 

\(\Rightarrow n=\left\{1;3\right\}\) 

\(b,\) Để \(n+8⋮n+1\)

\(\Rightarrow\left(n+1\right)+7⋮n+1\)

Mà \(n+1⋮n+1\Rightarrow7⋮n+1\)

\(\Rightarrow6⋮n\)

Mà n là số tự nhiên 

\(\Rightarrow n=\left\{1;2;3;6\right\}\)

+ 40xy chia hết cho 4 nên 40xy là số chẵn => y là số chẵn

+ 40xy chia hết cho 5 nên y=0 hoặc y=5 do y chẵn nên y=0

+ 40xy=40x0 chia hết cho 3 nên 4+x chia hết cho 3 nên x=2 hặc x=5 hoặc x=8

=> x={2,5,8}; y=0