\(A=\left(n+10\right)\left(n+15\right)\)

\(A=n^2+15n+10n+150\)

\(A=n^2+25n+150\)

Xét: 150 là 1 số chẵn.

Xét: Nếu n chẵn:

\(n^2;25n\) luôn chẵn

\(\Rightarrow n^2+25n+150\)= chẵn+chẵn+chẵn=chẵn \(⋮2\)

Xét: Nếu n lẻ:

\(\Rightarrow n^2;25n\) luôn lẻ

\(\Rightarrow n^2+25+150\)= lẻ+lẻ+chẵn=chẵn \(⋮2\)

\(\rightarrow A⋮2\rightarrowđpcm\)

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

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

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

\(B=3^2.3^{26}-3.3^{26}-3^{26}\)

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

\(B=3^{26}.5⋮5\)

\(B=\left(3^2\right)^{13}.5\)

\(B=9^{13}.5⋮9\)

\(B⋮5;9\Rightarrow B⋮45\rightarrowđpcm\)

Ta có:81⁷-27⁹-9¹³

       =(3⁴)⁷-(3³)⁹-(3²)¹³

       =3²⁸-3²⁷-3²⁶=3²⁶.(3²-3-1)=3²⁶.5=3²².3⁴.5=3²².405 luôn chia hết cho 405

=>đpcm

7^6+7^5-7^4 = 7^4*(7^2+7-1) = 7^4*55 

mình học lớp 5 mong bạn thông cảm và

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 

a)

\(3^{21}-3^{18}\\ =3^{17}.\left(3^4-3\right)\\ =3^{17}.\left(81-3\right)\\ =3^{17}.78\)

Vì \(3^{17}.78⋮78\) nên \(3^{21}-3^{18}⋮78\) (đpcm)

Vậy...

b)
\(81^7-27^9-9^{13}\\ =\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\\ =3^{28}-3^{27}-3^{26}\\ =3^{24}.\left(3^4-3^3-3^2\right)\\ =3^{24}.\left(81-27-9\right)\\ =3^{24}.45\)

Vì \(3^{24}.45⋮45\) nên \(81^7-27^9-9^{13}⋮45\) (đpcm)

Vậy...

81^7 trừ nha

c: \(81^7-27^9-9^{13}\)

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

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

\(=3^{24}\cdot45⋮45\)