![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27^8-3^{21}=\left(3^3\right)^8-3^{21}=3^{24}-3^{21}=3^{21}.\left(3^3-1\right)=3^{21}.26\) chia hết cho 26
\(8^{12}-2^{33}-2^{30}=8^{12}-\left(2^3\right)^{11}-\left(2^3\right)^{10}=8^{12}-8^{11}-8^{10}\)
\(=8^{10}.\left(8^2-8-1\right)=8^{10}.55\)
hình như đề sai ở câu sau nhé bn,chia hết cho 55 thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8^{15}-2^{43}+2^{41}=\left(2^3\right)^{15}-2^{43}+2^{41}\)
\(=2^{45}-2^{43}+2^{41}=2^{41}\left(16-4+1\right)=2^{41}.13⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=(1+4+4^2)+(4^3+4^4+4^5)+......+(4^2010+4^2011+4^2012)
=21+4^3.(1+4+4^2)+..........+4^2010.(1+4+4^2)
=21+4^3.21+...........+4^2010.21
=21.(1+4^3+....+4^2010) chia hết cho 21
Ta có \(1+4+4^2+4^3+....+4^{2012}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^6\right)+....+\left(4^{2010}+4^{2011}+4^{2012}\right)\)
\(=1.\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+....+4^{2010}.\left(1+4+4^2\right)\)
\(=\left(1+4+4^2\right).\left(1+4^3+....+4^{2010}\right)\)
\(=21.\left(1+4^3+...+4^{2010}\right)⋮21\)
Vậy biểu thức chia hết cho 21
![](https://rs.olm.vn/images/avt/0.png?1311)
=88-165
=224-220
=220.[24-1]
=220.15 chia hết cho 15
Vậy 88-165 chia hết cho 15
b,
=105-253
=55.25-56
=55.[25-5]
=55.27 chia hết cho 27
Vậy 105-253 chia hết cho 27
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(10^{19}+10^{18}+10^{17}=10^{16}.10^3+10^{16}.10^2+10^{16}.10=10^{16}.\left(1000+100+10\right)=10^{16}.1110\)
vì 1110 : 555 bằng 2
=> ................... chia hết cho 555
1) ( 1019+ 1018+1017) chia hết cho 555
= 1017.102+1018.10+1017
= 1017.(102+10+1)
= 1017.111
= 1016.10.111
= 1016.1110 = 1016.555.2
=> ( 1019+ 1018+1017) chia hết cho 555
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8\equiv1\left(mod7\right)\Rightarrow8^{13}\equiv1\left(mod7\right)\)
\(4^{20}=16.\left(4^3\right)^6=16.\left(64\right)^6=2.64^6+14.64^6\), mà \(64\equiv1\left(mod7\right)\Rightarrow2.64^3\equiv2\left(mod7\right)\)
\(\Rightarrow4^{20}\equiv2\left(mod7\right)\)
\(2^{41}=4.2^{39}=4.\left(2^3\right)^{13}=4.8^{13}\) , mà \(8\equiv1\left(mod7\right)\Rightarrow4.8^{13}\equiv4\left(mod7\right)\)
\(\Rightarrow8^{13}+4^{20}+2^{41}\equiv\left(1+2+4=7\right)\left(mod7\right)\)
Hay \(3^{13}+4^{20}+2^{41}⋮7\)
\(2^{41}+4^{21}+8^{15}=2^{41}+\left(2^2\right)^{21}+\left(2^3\right)^{15}=2^{41}+2^{42}+2^{45}=2^{41}\left(1+2+2^4\right)=2^{41}.19\) chia hết cho 19(đpcm)
Thách oOo KiRitO oOo tính kiểu đấy đấy