a,

ta có

\(12^{1980}-2^{1600}=\left( 12^4\right)^{495}-\left(2^4\right)^{400}=\left(...6\right)^{495}-\left(...6\right)^{400}=\left(...6\right)-\left(...6\right)=\left(...0\right)\)

có tận cùng bằng 0 nên \(\left(12^{1980}-2^{1600}\right)\)chia hết cho 10

                                                    Bài giải

\(a,\text{ }12^{1980}-2^{1600}=\left(3\cdot2^2\right)^{1980}-\left(2^4\right)^{400}=3^{1980}\cdot2^{3960}-216^{400}\)

\(=\left(3^4\right)^{495}\cdot\left(2^4\right)^{990}-216^{40}=\overline{\left(...1\right)}^{495}\cdot\overline{\left(...6\right)}^{990}-\overline{\left(...6\right)}^{495}=\overline{\left(...1\right)}\cdot\overline{\left(...6\right)}-\overline{\left(...6\right)}\)

\(=\overline{\left(...6\right)}-\overline{\left(...6\right)}=\overline{\left(...0\right)}\text{ }\)

Vì số có chữ số tận cùng là 0 thì chia hết cho 10 \(\Rightarrow\text{ }\left(12^{1980}-2^{1600}\right)\text{ }⋮\text{ }10\)

                                                                       Bài giải

\(b,\text{ }\left(19^{2005}-11^{2006}\right)=19^{2004}\cdot19-11^{2006}=\left(19^2\right)^{1002}\cdot19-11^{2006}\)

\(=\overline{\left(...1\right)}^{1002}\cdot19-\overline{\left(...1\right)}^{2006}=\overline{\left(...9\right)}-\overline{\left(...1\right)}=\overline{\left(...8\right)}\text{ }⋮̸\text{ }10\)

\(\Rightarrow\text{ Vô lí}\)

a, \(12^{1980}-2^{1600}\)

\(=\left(2^4\right)^{495}-\left(2^4\right)^{400}\)

\(=16^{495}-16^{400}\)

\(=\overline{...6}-\overline{...6}\)

\(=\overline{...0}⋮10\left(đpcm\right)\)

b, \(19^{2005}+11^{2006}\)

\(=19\cdot19^{2004}+\overline{...1}\)

\(=19\cdot\left(19^2\right)^{1002}+\overline{...1}\)

\(=19\cdot361^{1002}+\overline{...1}\)

\(=19\cdot\overline{...1}+\overline{...1}\)

\(=\overline{...9}+\overline{...1}\)

\(=\overline{...0}⋮10\left(đpcm\right)\)

(đpcm) là j vậy bạn

ta có

12^1980=(12^4)^495=20736^495=(.....6)

2^1600=(2^4)^400=16^400=(.....6)

=>12^1980-2^1600=(...6)-(....6)=(...0)chia hết cho 10(đpcm)

Vậy...