Question

chứng minh rằng

a) \(8^7-2^{18}\) chia hết cho 14

b) \(5^5-5^4+5^3\) chia hết cho 7

c) \(7^6+7^5-7^4\) chia hết cho 11

bn nào bt lm lm giúp mk vs mai mk nộp r

Answer 1

\(a.\)

\(8^7-2^{18}\)

\(=\left(2^3\right)^7-2^{18}\)

\(=2^{21}-2^{18}\)

\(=2^{18}.2^3-2^{18}\)

\(=2^{18}\left(2^3-1\right)\)

\(=2^{18}.7\)

\(=2^{17}.7.2⋮14\)

Vậy \(8^7-2^{18}⋮14\)

\(b.\)

\(5^5-5^4+5^3\)

\(=5^3\left(5^2-5+1\right)\)

\(=5^3.21\)

\(=5^3.7.3⋮7\)

Vậy \(5^5-5^4+5^3⋮7\)

\(c.\)

\(7^6+7^5-7^4\)

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

\(=7^4.55\)

\(=7^4.5.11⋮11\)

Vậy \(7^6+7^5-7^4⋮11\)

Answer 2

mk chỉ bt làm phần b với c thui xin lỗi bn nha

Answer 3

b, ta có:5^5-5^4+5^3

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

=5^3.21

=5^3.3.7 chia hết cho 7

c,ta có: 7^6+7^5-7^4

=7^4(7^2+7^1-1)

=7^4.55

=7^4.5.11 chia hết cho 11