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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=7^3+7^4+....+7^{98}\)
\(\Rightarrow A=7^3\left(1+7\right)+......+7^{97}\left(1+7\right)\)
\(\Rightarrow A=7^3.8+......+7^{97}.8\)
=> A chia hết cho 8
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=7^3+7^4+7^5+7^6+...+7^{98}\)
\(\Rightarrow A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...+\left(7^{97}+7^{98}\right)\)
\(\Rightarrow A=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\)
\(\Rightarrow A=7^3.8+7^5.8+...+7^{97}.8\)
\(\Rightarrow A=\left(7^3+7^5+...+7^{97}\right).8⋮8\)
\(\Rightarrow A⋮8\)
Vậy \(A⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=7^3+7^4+7^5+...+7^{97}+7^{98}\)
\(=\left(7^3+7^4\right)+\left(7^5+7^6\right)+\left(7^7+7^8\right)+...+\left(7^{97}+7^{98}\right)\)
\(=7^3\left(1+7\right)+7^5\left(1+7\right)+7^7\left(1+7\right)+...+7^{97}\left(1+7\right)\)
\(=8\left(7^3+7^5+7^7+...+7^{97}\right)\) \(⋮8\) (đpcm)
Ta có :
A = 73 + 74 + 75 + 76 +........+797 + 798
=> A = 73 ( 1+ 7)+...........+797 ( 1+7)
=> A = 73 x 8 +.......+798 x 8
=> A chia hết cho 8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=7^3+7^4+7^5+7^6+...+7^{97}+7^{98}\)
\(=\left(7^3+7^4\right)+\left(7^5+7^6\right)+....+\left(7^{97}+7^{98}\right)\)
\(=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\)
\(=\left(1+7\right)\left(7^3+7^5+...+7^{97}\right)\)
\(=8\left(7^3+7^5+...+7^{97}\right)⋮8\)
Vì A có: 96 số hạng nên ta chia A thành 48 nhóm 1 nhóm có 2 số hạng
\(A=7^3+7^4+7^5+7^6+...........+7^{97}+7^{98}\)
\(A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...........+\left(7^{97}+7^{98}\right)=7^3\left(1+7\right)+7^5\left(1+7\right).....+7^{97}\left(1+7^{ }\right)\)
\(A=7^3.8+7^5.8+.......+7^{97}.8=8\left(7^3+7^5+........+7^{97}\right)⋮8\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(A=7^3+7^4+7^5+...+7^{97}+7^{98}.\)
\(A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...+\left(7^{97}+7^{98}\right).\)
\(A=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right).\)
\(A=7^3.8+7^5.8+...+7^{97}.8.\)
\(A=\left(7^3+7^5+...+7^{97}\right).8⋮8\left(đpcm\right).\)
Vậy.....
Bài 2: Tìm x:
\(3^{47}:\left(189-3x\right)=3^{44}.\)
\(189-3x=3^{47}:3^{44}.\)
\(189-3x=27.\)
\(3x=189-27.\)
\(3x=162.\)
\(x=162:3.\)
\(x=54.\)
Vậy.....
\(27-3.\left(5x+2\right)=6.\)
\(3.\left(5x+2\right)=27-6.\)
\(3.\left(5x+2\right)=21.\)
\(5x+2=21:3.\)
\(5x+2=7.\)
\(5x=7-2.\)
\(5x=5.\)
\(x=5:5.\)
\(x=1.\)
Vậy.....
\(A=7^3+7^4+...+7^{98}\\ \Rightarrow A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+....+\left(7^{97}+7^{98}\right)\\ =7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\\ =7^3.8+7^5.8+...+7^{97}.8\\ =8\left(7^3+7^5+...+7^{97}\right)⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(7 mũ 3 + 7 mũ 4)+.............(7 MŨ 97+7 mũ 98)
A=7 mũ 3 x(7+1)+............7 mũ 97 x (7+1)
A=7 mũ 3 x8.......................7 mũ 97 x8
A=8 x(7 mũ 3 +.............7 mũ 97)
vì ta thấy 8 chia hết cho 8 nên tổng trên chi hết cho 8
tsch mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 73 + 74 + 75 + ... + 797 + 798
A = ( 73 + 74 ) + ( 75 + 76 ) + ... + ( 797 + 798 )
A = 73 . ( 1 + 7 ) + 75 . ( 1 + 7 ) + ... + 797 . ( 1 + 7 )
A = 73 . 8 + 75 . 8 + ... + 797 . 8
A = 8 . ( 73 + 75 + ... + 797 ) \(⋮8\)
Vậy A chia hết cho 8 ( dpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(5+5^2)+(5^3+5^4)+...(5^299+5^300)
A=5(1+5)+5^2(1+5)+...+5^299(1+5)
A=5.6+5^2.6+...+5^299.6 => Achia hết cho 6.
Tường tự phần A nhóm 3 số với nhau chia hết cho 31
phần B đường nhiên sẽ chia hết cho 7 vì mỗi số hạng đều chia hết cho 7, nhóm 2 số với nhau chia hết cho 8
![](https://rs.olm.vn/images/avt/0.png?1311)
A=72+74+76+78
A=72.(72+1)+76(1+72)
A=50.(72+76)\(⋮\)5
→A\(⋮\)5
A = 73 + 74 + 75 + 76 + ... + 797 + 798
A = ( 73 + 74 ) + ( 75 + 76 ) + .... + ( 797 + 798 )
A = 73 . ( 1 + 7 ) + 75 . ( 1 + 7 ) + ... + 797 . ( 1 + 7 )
A = 73 . 8 + 75 . 8 + .... + 797 . 8
A= 8 . ( 73 + 75 + ..... + 797 ) \(⋮8\)
Vậy A \(⋮8\)( dpcm )