Ta có :

\(5+5^2+5^3+5^4+....+5^8\)

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

\(=30+30.5^2+...+30.5^6\)

\(=30.\left(1+5^2+...+5^6\right)\)

\(=3.10.\left(1+5^2+...+5^6\right)⋮3\)

Vậy \(5+5^2+5^3+5^4+...+5^8\)chia hết cho 3 .

Học tốt

                   Bài làm :

Ta có :

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

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

\(=5\left(1+5\right)+5^3\left(1+5\right)+5^5\left(1+5\right)+5^7\left(1+5\right)\)

\(=\left(1+5\right)\left(5+5^3+5^5+5^7\right)\)

\(=6.\left(5+5^3+5^5+5^7\right)\)

Vì 6 chia hết cho 3

\(\Rightarrow6.\left(5+5^3+5^5+5^7\right)⋮3\)

=> Điều phải chứng minh

1:\(A=1+3+3^2+3^3+...+3^{11}\)

\(A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\)

\(A=4+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\)

\(A=4+3^2\cdot4+....+3^{10}\cdot4\)

\(A=4\cdot\left(1+3^2+...+3^{10}\right)\) chia hết cho 4

Vì ta có 4 chia hết cho 4 => A có chia hết cho 4

Vậy A chia hết cho 4

2:

\(C=5+5^2+5^3+...+5^8\) chia hết cho 30

\(C=\left(5+5^2\right)+...+\left(5^7+5^8\right)\)

\(C=30+5^2\cdot\left(5+5^2\right)+...+5^6\cdot\left(5+5^2\right)\)

\(C=30\cdot1+5^2\cdot30+...5^6\cdot30\)

\(C=30\cdot\left(5^2+...+5^6\right)\)

Vì ta có 30 chia hết cho 30 nên suy ra C có chia hết cho 30

Vậy C có chia hết cho 30

4^3 * 32^5 - 8^8 k chia hết cho 5

b: \(8^{10}-8^9-8^8=8^8\left(8^2-8-1\right)=8^8\cdot55⋮55\)

c: 5^5-5^4+5^3

=5^3(5^2-5+1)

=5^3*21 chia hết cho 7

e:

72^63=(3^2*2^3)^63=3^126*2^189

 \(24^{54}\cdot54^{24}\cdot10^2=2^{162}\cdot3^{54}\cdot3^{72}\cdot2^{24}\cdot2^2\cdot5^2\)

\(=2^{188}\cdot3^{136}\cdot5^2\) chia hết cho 3^126*2^189

=>ĐPCM

g: \(=\left(3^4\right)^7-\left(3^3\right)^9-3^{26}\)

\(=3^{26}\left(3^2-3-1\right)=5\cdot3^{26}=5\cdot9\cdot3^{24}⋮5\cdot9=45\)

\(5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9\)

\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)

\(=5\times\left(1+5+5^2\right)+5^4\times\left(1+5+5^2\right)+5^7\times\left(1+5+5^2\right)\)

\(=5\times31+5^4\times31+5^7\times31\)

\(=31\times\left(5+5^4+5^7\right)⋮31\)

Vậy tổng trên chia hết cho 31

            Bài làm :

Ta có :

\(5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9\)

\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)

\(=5\times\left(1+5+5^2\right)+5^4\times\left(1+5+5^2\right)+5^7\times\left(1+5+5^2\right)\)

\(=5\times31+5^4\times31+5^7\times31\)

\(=31\times\left(5+5^4+5^7\right)⋮31\)

=> Điều phải chứng minh

Ta có:

Do \(2^2>1.2\) ; \(3^2>2.3\) ;...; \(9^2>8.9\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

\(\Rightarrow A< 1-\dfrac{1}{9}< 1\) (1)

Lại có: \(2^2< 2.3\) ; \(3^2< 3.4\) ;...; \(9^2< 9.10\)

\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{2}{5}\) (2)

(1);(2) \(\Rightarrow\dfrac{2}{5}< A< 1\)

Ta có A = \(5+5^2+5^3+...+5^8\)

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

= 30 + \(5^2\left(5+5^2\right)+....+5^6\left(5+5^2\right)\)

= \(30+5^2.30+5^3.30+...+5^6.30\)

= \(30\left(1+5^2+5^3+5^4+5^5+5^6\right)\) chia hết cho 30

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

\(=1\left(5+25\right)+5^2\left(5+25\right)+5^4\left(5+25\right)+5^6\left(5+25\right)\)

\(=1.30+5^2.30+5^4.30+5^6.30\)

\(=30\left(1+5^2+5^4+5^6\right)⋮30\)  (đpcm)