#)Giải :

Ta có : \(\left(81^7-27^9-9^{13}\right)\)

\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

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

\(=3^{26}.3^2-3^{26}.3-3^{26}\)

\(=3^{26}\left(3^2-3-1\right)\)

\(=3^{26}.5\)

\(=3^{22}.3^4.5\)

\(=3^{22}.405\)chia hết cho 405 ( đpcm )

Sửa đề: Chứng minh cái biểu thức trên chia hết cho 405.

Thật vậy,xét theo mod405:

\(81^7\equiv81^5.81^2\equiv81.81^2\equiv81\left(mod405\right)\)

\(27^9\equiv27^5.27^4\equiv162.81\equiv162\left(mod405\right)\)

\(9^{13}\equiv9^7.9^6\equiv324.81\equiv324\)

Suy ra \(81^7-27^9-9^{13}\equiv81-162-324\equiv-405\equiv0\left(mod405\right)\)

Hay ta có đpcm.

\(81^7-27^9-9^{13}\)

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

\(=3^{26}\left(3^2-3-1\right)\)

\(=3^{24}\cdot9\cdot5⋮45\)

\(\Rightarrow3^{28}-3^{27}-3^{26}=3^{26}.\left(3^2-3-1\right)=3^{26}.5=3^{24}.9.5=3^{24}.45⋮45\)

a, 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) 
=3^26.5=3^13.3^2.5=45.3^13 chia hết cho 45 

\(81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26} \)

=\(3^{26}.(3^2-3-1)=3^{26}.5=9.5.3^{24}=45.3^{24}\)

Kết luận nữa là ok rồi bạn!!!

Chứng minh rằng:\(81^7-27^9-9^{13}\) chia hết cho 45

Giải:Ta có:\(81^7-27^9-9^{13}\)

= (3⁴)⁷ - (3³)⁹ - (3²)¹³

= 3²⁸ - 3²⁷ - 3²⁶ = 3²⁶.(3²-3-1)

= 3²⁶. (9 - 3 - 1) = 3²⁶.5 = 3²⁴.3².5=3²⁴.45 chia hết cho 45

Ta có:\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)

\(\Rightarrow3^4.3^{24}-3^3.3^{24}-3^2-3^{24}=\left(3^4-3^3-3^2\right).3^{24}=\left(81-27-9\right).3^{24}=45.3^{24}⋮45\)

Vậy \(81^7-27^9-9^{13}⋮45\)