b) Để 4x + 19 chia hết cho x + 1 thì 15 chia hết cho x + 1

--> x + 1 là ước của 15

TH1: x + 1 = 15 <=> x = 14

TH2: x + 1 = 1 <=> x = 0

TH3: x + 1 = 3 <=> x = 2

TH4: x + 1 = 5 <=> x= 4

\(7^1+7^2+7^3+...+7^{117}+7^{118}=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{116}\left(1+7+7^2\right)\)

\(=7.57+7^4.57+...+7^{116}.57=57\left(7+7^4+...+7^{116}\right)⋮57\)

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:

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

đúng rùi

a: \(B=3^1+3^2+...+3^{2010}\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{2009}\right)⋮4\)

\(B=3\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2008}\right)⋮13\)

b: \(C=5^1+5^2+...+5^{2010}\)

\(=5\left(1+5\right)+...+5^{2009}\left(1+5\right)\)

\(=6\left(5+...+5^{2009}\right)⋮6\)

\(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)\)

\(=31\left(5+...+5^{2008}\right)⋮31\)

c: \(D=7\left(1+7\right)+...+7^{2009}\left(1+7\right)\)

\(=8\left(7+...+7^{2009}\right)⋮8\)

\(D=7\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{2008}\right)⋮57\)

Ta có :

A chia hết cho 8 vì mọi số hạng của A deduf chia hết cho 8 .

\(A=8+2^2+....+8^{2019}\)

\(\Rightarrow A=8\left(1+8\right)+.....+8^{2018}\left(1+8\right)\)

\(\Rightarrow A=8.9+.....+8^{2018}.9\)

=> A chia hết cho 9 .

Mà (8;9)=1

=> A chia hết cho 8x9=72

\(A=8\left(1+8+8^2\right)+....+8^{2017}\left(1+8+8^2\right)\)

\(A=8.73+....+8^{2017}.73\)

=> A chia hết cho 73

Các bạn trả lời gấp giúp mình nhá!!!

9h mình phải đi học rùi

A = 8 + 8^2 +8^3 +...+ 8^58+8^59+8^60

   = (8+8^2 + 8^3) +...+ (8^58+8^59 +8^60)

   =8( 1+8+8^2)+...+8^58(1+8+8^2)

   = 8. 73 + ......+8^58 .73

   = 73.( 8+...+8^58) chia hết cho 73

Có:

+) \(81^4\equiv60\left(mod71\right)\)

\(\left(81^4\right)^2\equiv60^2\equiv50\left(mod71\right)\) (1)

+) \(27^5\equiv20\left(mod71\right)\)

\(\left(27^5\right)^2\equiv20^2\equiv45\left(mod71\right)\) (2)

+) \(9^7\equiv54\left(mod71\right)\)

\(\left(9^7\right)^2\equiv54^2\equiv5\left(mod71\right)\) (3)

Từ (1), (2), (3):

\(\Rightarrow81^8-27^{10}-9^{14}\equiv50-45-5\equiv0\left(mod71\right)\)

=> \(81^8-27^{10}-9^{14}⋮71\left(đpcm\right)\)