Question

I. chứng minh rằng

a. \(7^6+7^5-7^4⋮11\)

b. \(81^7-27^9-9^{13}⋮45\)

Answer 1

Lời giải:

a) Ta có:

\(7^6+7^5-7^4=7^{4+2}+7^{4+1}-7^4\)

\(=7^4.7^2+7^4.7-7^4=7^4(7^2+7-1)=7^4.55=11.7^4.5\vdots 11\) (đpcm)

b)

\(81^7-27^9-9^{13}=(3^4)^7-(3^3)^9-(3^2)^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=3^{26}(3^2-3-1)=5.3^{26}=5.3.3.3^{24}=45.3^{24}\vdots 45\) (đpcm)

Answer 2

a, \(7^6+7^5-7^4⋮11\)

= \(7^4.7^2+7^4.7-7^4\)

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

= \(7^4.\left(49+7-1\right)\)

=\(7^4.55=7^4.5.11\) => chia hết cho 11

b, \(81^7\)- \(27^9\)- \(9^{13}\)
=\(\left(3^4\right)^7\)- \(\left(3^3\right)^9\) - \(\left(3^2\right)^{13}\)
= \(3^{28}-3^{27}-3^{26}\)
=\(3^{26}.\left(3^2-3-1\right)\)
=3^26.5=3^13.3^2.5=45.3^13 chia hết cho 45