5^2005+5^2003

= 5^2003(5^2+1)

= 5^2003.26

=5^2003.13.2

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

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

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

\(=5^{2003}.2.13\)\(⋮\)\(13\)

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

                         = 5^2003 .26 chia hết cho 13

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

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

CHỨNG MINH LÊ NHẬT PHÚC NGU NGƯỜI

ko biết

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

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...

a) 5+5²+5³+5⁴+...+5¹⁰⁰

= (5+5²)+(5³+5⁴)+...+(5⁹⁹+5¹⁰⁰)

= 30+5².(5+5²)+...+5⁹⁸.(5+5²)

= 30+5².30+...+5⁹⁸.30

= 30.(1+5²+...+5⁹⁸)

Vì 30 chia hết cho 10

=> 30.(1+5²+...+5⁹⁸) chia hết cho 10

=> 5+5²+5³+...+5¹⁰⁰ chia hết cho 10