5^2005+5^2003

= 5^2003(5^2+1)

= 5^2003.26

=5^2003.13.2

a) Ta có:

\(n\left(2n-3\right)-2n\left(n+1\right)\)

\(=2n^2-3n-2n^2-2n\)

\(=-5n\)

Vì \(-5n⋮5\) với n thuộc Z

\(\Rightarrow n\left(2n-3\right)-2n\left(n+1\right)⋮5\) với n thuộc Z

b) Ta có:

\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)

\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)

\(=5n^2+5n\)

\(=5\left(n^2+n\right)\)

Vì \(5\left(n^2+n\right)⋮5\)

\(\Rightarrow\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)

c) Ta có:

\(\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)

\(=\left(xy+1-2\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)

\(=\left(xy+1\right)\left(x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)

\(=\left(xy+1\right)\left(x^{2003}+y^{2003}-x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)\)

\(=2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)\)

Vì \(2\left(xy+1\right)y^{2003}⋮2\)

\(2\left(x^{2003}+y^{2003}\right)⋮2\)

\(\Rightarrow2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)⋮2\)

\(\Rightarrow\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)⋮2\)

\(5^{2005}+5^{2003}\)

\(=5^{2003}.\left(5^2+1\right)\)

\(=5^{2003}.26\)

\(=5^{2003}.2.13\)\(⋮\)\(13\)

5^2005 + 5^2003 = 5^2003 (5^2 +1)

                         = 5^2003 .26 chia hết cho 13

\(5^{2005}+5^{2003}=5^{2003}\left(5^2+1\right)=5^{2003}.26=5^{2003}.13.2⋮13\)

\(5^{2003}+5^{2002}+5^{2001}\)

\(=5^{2001}\left(5^2+5+1\right)\)

\(=5^{2001}.31\)chia hết cho 31.

a) Vì abcd chia hết cho 4 nên 10c + d chia hết cho 4

Mặt khác 10c + d = 8c + 2c + d

Vì 8c chia hết cho 4 nên 2c + d cũng chia hết cho 4
 

Có 13 giao thừa = 1.2.3.4.5.6.7.8.9.10.11.12.13 chia hết cho 2

Có 11 giao thừa = 1.2.3.4.5.6.7.8.9.10.11 chia hết cho 2

suy ra 13 giao thừa - 11 giao thừa chia hết cho 2

xin các bạn k cho mình nhé