5²⁰⁰⁵+5²⁰⁰³
=5²⁰⁰³.(5²+1) 
=5²⁰⁰³.26 
=5²⁰⁰³.13.2 

Vì 13 chia hết cho 13 nên 5²⁰⁰³ . 13 . 2 chia hết 13

Vậy: 5²⁰⁰⁵+5²⁰⁰³

Ta có: \(5^{2005}+5^{2003}=5^{2003}\left(5^2+1\right)=5^{2003}.26=5^{2003}.2.13\)  chia hết cho  \(13\)

Vậy,  \(5^{2005}+5^{2003}\)  chia hết cho  \(13\)

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

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

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

\(=5^{2003}.2.13\)\(⋮\)\(13\)

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

                         = 5^2003 .26 chia hết cho 13

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

a) Ta có:

\(5^2=25\equiv-1\left(mod13\right)\)

\(\Rightarrow\left\{{}\begin{matrix}5^{2004}=\left(5^2\right)^{1002}\equiv\left(-1\right)^{1002}\left(mod13\right)\equiv1\left(mod13\right)\\5^{2002}=\left(5^2\right)^{1001}\equiv\left(-1\right)^{1001}\left(mod13\right)\equiv-1\left(mod13\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5^{2005}=5^{2004}.5\equiv1.5\left(mod13\right)\equiv5\left(mod13\right)\\5^{2003}=5^{2002}.5\equiv\left(-1\right).5\left(mod13\right)\equiv-5\left(mod13\right)\end{matrix}\right.\)

\(\Rightarrow5^{2005}+5^{2003}\equiv5+\left(-5\right)\left(mod13\right)\equiv0\left(mod13\right)\)

Vậy...

Bài 1:

a,\(5^{2005}+5^{2003}=5^{2003}(25+1)=26.5^{2003}\vdots13(đpcm)\)

b,\(a^2+b^2+1\ge ab+a+b\)

<=>\(2a^2+2b^2+2\ge2ab+2a+2b\)

<=>\((a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)\ge0\)

<=>\((a-b)^2+(a-1)^2+(b-1)^2\ge0(tm)\)

=> đpcm

a) 5²⁰⁰⁵ + 5²⁰⁰³ = 5²⁰⁰³ ( 5² + 1 ) = 5²⁰⁰³ . 26 = 5²⁰⁰³ . 2 .13

=> 5²⁰⁰⁵ + 5²⁰⁰³ chia hết cho 13

b) a² + b² +1 \(\ge\) ab + a + b

\(\Leftrightarrow\) 2a² + 2b² + 2 ≥ 2ab + 2a + 2b

\(\Leftrightarrow\)(a² − 2ab + b²) + (a² − 2a + 1) + (b² − 2b + 1) ≥ 0

\(\Leftrightarrow\) (a − b)² + (a − 1)² + (b − 1)² ≥ 0

a)2004¹⁰⁰+2004⁹⁹=2004⁹⁹(2004+1)=2014⁹⁹.2005

=>2014⁹⁹.2005chia hết cho 2005

=> 2004¹⁰⁰+2004⁹⁹ chia hết cho 2005

b) 4¹³+32⁵-8⁸

=(2²)¹³+(2⁵)⁵-(2³)⁸

=2²⁶+2²⁵-2²⁴

=2²⁴(2²+2-1)

=2²⁵.5

=>2²⁵chia hết cho 5 => 4¹³+32⁵-8⁸ chia hết cho 5

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 \(2\left(xy+1\right)y^{2003}⋮2\)

\(2\left(x^{2003}+y^{2003}\right)⋮2\)

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

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