Question

bài 1: cmr

a) 5²⁰⁰⁵ + 5²⁰⁰³ chia hết cho 13

b) a² + b² +1 \(\ge\) ab + a + b

Answer 1

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

Answer 2

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