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\(8^5+16^4=\left(2^3\right)^5+\left(2^4\right)^4=2^{15}+2^{16}=2^{15}.1+2^{15}.2=2^{15}\left(2+1\right)=2^{15}.3\)
Vậy tổng chia hết cho 3
\(2^8+2^9+2^{10}=2^8.1+2^8.2+2^8.2^2=2^8.\left(1+2+4\right)=2^8.7\)
Vậy tổng chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4n+3\right)^2-25=\left(4n+3-5\right)\left(4n+3+5\right)\)
\(=\left(4n-2\right)\left(4n+8\right)=2.\left(2n-1\right).4.\left(n+2\right)=8\left(2n-1\right)\left(n+2\right)⋮8\)
\(\left(2n+3\right)^2-9=\left(2n+3-3\right)\left(2n+3+3\right)\)
\(=2n\left(2n+6\right)=4n\left(n+3\right)⋮4\)
\(\left(3n+4\right)^2-16=\left(3n+4-4\right)\left(3n+4+4\right)\)
\(=3n\left(3n+8\right)⋮3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(8^5+16^4\)
\(=\left(2^3\right)^5+\left(2^4\right)^4\)
\(=2^{15}+2^{16}\)
\(=2^{15}\left(1+2\right)\)
\(=2^{15}.3\)chia hết hết cho 3.
b) \(2^8+2^9+2^{10}\)
\(=2^8\left(2^2+2+1\right)\)
\(=2^8.7\)chia hết cho 7
chứng minh : a) 8^5+16^4 chia hết cho 3
b) 2^8+2^9+2^10 chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\left(4n-7-5\right)\left(4n-7+5\right)\)
\(=\left(4n-12\right)\left(4n-2\right)\)
\(=8\left(n-3\right)\left(2n-1\right)⋮8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^8+2^6+2^{10}=2^6\left(2^2+1+2^4\right)=2^6.21=2^6.3.7;\) chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a)251-1
=(23)17-1\(⋮\)23-1=7
Vậy 251-1\(⋮\)7
b)270+370
=(22)35+(32)35\(⋮\)22+32=13
Vậy 270+370\(⋮\)13
c)1719+1917
=(BS18-1)19+(BS18+1)17
=BS18-1+BS18+1
=BS18\(⋮\)18
d)3663-1\(⋮\)35\(⋮\)7
Vậy 3663-1\(⋮\)7
3663-1
=3663+1-2
=BS37-2\(⋮̸\)37
Vậy 3663-1\(⋮̸\)37
e)24n-1
=(24)n-1\(⋮\)24-1=15
Vậy 24n-1\(⋮\)15
\(2^8+2^9+2^{16}=2^8\left(1+2+2^8\right)=2^8\cdot259=2^8\cdot7\cdot37⋮7\)
Ta có :
28 + 29+ 216 = 28+9+16 = 233=8589934592
Mà khi phân tích dựa theo 1 số \(⋮\)cho 7 nên 8589934592 \(⋮\)cho 7.