a) \(2010^{100}+2010^{99}\)

\(=2010^{99}\left(2010+1\right)\)

\(=2010^{99}.2011⋮2011\left(dpcm\right)\)

b) \(3^{1994}+3^{1993}-3^{1992}\)

\(=3^{1992}\left(3^2+3-1\right)\)

\(=3^{1992}.11⋮11\left(dpcm\right)\)

c) \(4^{13}+32^5-8^8\)

\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)

\(=2^{26}+2^{25}-2^{24}\)

\(=2^{24}\left(2^2+2-1\right)\)

\(=2^{24}.5⋮5\left(dpcm\right)\)

a.2014¹⁰⁰  + 2014⁹⁹

=2014⁹⁹.(2014+1)

=2014⁹⁹.2015

=> 2014¹⁰⁰  + 2014⁹⁹ chia hết cho 2015

 b.3¹⁹⁹⁴ + 3^1993  _  3¹⁹⁹² 

=3¹⁹⁹².(3²+3-1)

=3¹⁹⁹².11

=>3¹⁹⁹⁴ + 3^1993  _  3¹⁹⁹² chia hết cho 11

c. 4^{13 _} 32^{5 _} 8⁸

=(2²)¹³-(2⁵)⁵-(2³)⁸

=2²⁶-2²⁵-2²⁴

=2²⁴.(2²-2-1)

=2²⁴.5

=> 4^{13 _} 32^{5 _} 8⁸ chia hết cho 5

a)\(2014^{100}+2014^{99}=2014^{99}.\left(2014+1\right)=2014^{99}.2015⋮2015\left(\text{Đ}PCM\right)\)

b)\(3^{1994}+3^{1993}-3^{1992}=3^{1992}.\left(3^2+3-1\right)=3^{1992}.\left(9+3-1\right)=3^{1992}.11⋮11\left(\text{Đ}PCM\right)\)

c)\(4^{13}-32^5-8^8=\left(2^2\right)^{13}-\left(2^5\right)^5-\left(2^3\right)^8=2^{26}-2^{25}-2^{24}=2^{24}.\left(2^2-2-1\right)\)

Đề sai rồi bạn 2^14 luôn tận cùng chẵn =>2^14 không chia hết cho 5

Chúc bạn học tốt

1) 3^1994+4^1993-3^1992

  = 3^1992.(9+3-1)=3^1992.11 chia hết cho 11

=> 3^1994+3^1993-3^1992 chia hết cho 11

Có ai bt bài 2 ko z 

B2 

a)2^n =4.32=128=2^7

=>n=7

b)3^3n.3^2n =9^25

=>3^5n=3^50 =>n=10

Phạm Hồng Phúc lấy máy tính mà tính á bn