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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^6\left(5+5^2\right)=30+5^2.30+...+5^6.30\)
\(=30\left(1+5^2+...+5^6\right)⋮30\Rightarrowđpcm\)
b) \(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)=273+3^6.273+...+3^{24}.273\)
\(=273.\left(1+3^6+...+3^{24}\right)⋮273\Rightarrowđpcm\)
a: \(B=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)\)
\(=156\cdot5\cdot\left(1+5^4\right)\)
\(=780\left(1+5^4\right)⋮30\)
b: \(B=\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^2+3^5\right)\)
\(=273\cdot\left(1+...+3^{24}\right)⋮273\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=5+5^2+5^3+........+5^8\)
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+......+\left(5^7+5^8\right)\)
\(A=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^7.\left(1+5\right)\)
\(A=5.6+5^3.6+......+5^7.6\)
\(A=6.\left(5+5^3+.....+5^7\right)\)
\(A=3.2.\left(5+5^3+....+5^7\right)\)
\(\Leftrightarrow A⋮2\)
mà \(A⋮5\)
\(\Rightarrow A⋮\left(2.5\right)\Leftrightarrow A⋮10\)
Vậy A là bội của 10.
b/ \(B=3+3^3+3^5+...+3^{39}\)
\(B=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+....+\left(3^{35}+3^{37}+3^{39}\right)\)
\(B=1.\left(3+3^3+3^5\right)+3^6.\left(3+3^3+3^5\right)+....+3^{34}.\left(3+3^3+3^5\right)\)
\(B=1.273+3^6.273+.....+3^{34}.273\)
\(B=273.\left(1+3^6+...+3^{34}\right)\)
\(\Rightarrow B⋮273\)
Vậy B là bội của 273
Nhớ k cho mình nhé! Thank you!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+.............+3^{24}\left(3+2^3+3^5\right)\)
\(B=273+273\cdot3^6+.............+273\cdot3^{24}\)
\(B=273\left(1+3^6+.......+3^{24}\right)⋮273\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(B=3+3^3+3^5+...+3^{29}\)
\(\Rightarrow B=\left(3+3^3+3^5\right)+...+\left(3^{25}+3^{27}+3^{29}\right)\)
\(\Rightarrow B=\left(3+3^3+3^5\right)+...+3^{24}.\left(3+3^3+3^5\right)\)
\(\Rightarrow B=273+...+3^{24}.273\)
\(\Rightarrow B=273.\left(1+...+3^{24}\right)⋮273\)
Vậy B là bội của 273.
b) \(A=5+5^2+...+5^7+5^8\)
\(\Rightarrow A=\left(5+5^2\right)+...+\left(5^7+5^8\right)\)
\(\Rightarrow A=\left(5+5^2\right)+...+5^6.\left(5+5^2\right)\)
\(\Rightarrow A=30+...+5^6.30\)
\(\Rightarrow A=30.\left(1+...+5^6\right)⋮30\)
Vậy A là bội của 30.
A=5+5^2+5^3+...+5^20
=(5+5^2)+(5^3+5^4)+...+(5^19+5^20)
=(5+5^2)+5^2(5+5^2)+...5^18(5+5^2)
=30+5^2.30+5^4.30+5^6.30+..+5^18.30
=30(1+5^2+5^4+5^6+..+5^18)(chia hết cho 30)
Vậy A là bội của 30
a, đề phải là cm ko chia hết cho 5
A = 5+5^2+(5^3+5^4)+(5^5+5^6)+(5^7+5^8)
= 30 + 5.(5^2+5^3)+5^3.(5^2+5^3)+5^5.(5^2+5^3)
= 30+5.150+5^3.150+5^5.150
= 30+150.(5+5^3+5^5)
Vì 150 chia hết cho 50 => 150.(5+5^3+5^5) chia hết cho 50
Mà 30 ko chia hết cho 50
=> A ko chia hết cho 50