Nó có chia hết à ??? 

Ta thấy 2009 chia 2010 dư  -1 

=> 2009 ^ 2008 chia 2010 dư 1 (1)

Lại có  2011 chia 2010 dư 1

=> 2011^2010 chia 2020 dư 1 (2)

Từ (1)(2) => 2009^2008-2011^2020 chia 2010 dư 2 (sai )

2009^2008+2011^2010 chia hết cho 2010 2009^2008+2011^2010

=2009^2008+2011^2010

=2009^2008+2011^2010+1-1

=(2009^2008+ 1) + (2011^2010– 1)

= (2009 + 1)(2009^2007- …) + (2011 – 1)(2011^2009 + …)

= 2010(2009^2008 - …) + 2010(2011^2009+ …) chia hết cho 2010  

2009^2008+2011^2010 chia hết cho 2010 2009^2008+2011^2010

=2009^2008+2011^2010

=2009^2008+2011^2010+1-1

=(2009^2008+ 1) + (2011^2010– 1)

= (2009 + 1)(2009^2007- …) + (2011 – 1)(2011^2009 + …)

= 2010(2009^2008 - …) + 2010(2011^2009+ …) chia hết cho 2010  

From: exoplanet

To: Nguyễn Ngọc Phương Thảo

\(2009^{2008}+2011^{2010}=\left(2009^{2008}+1\right)+\left(2011^{2010}-1\right)\)

\(=\left(2009+1\right)\left(2009^{2007}+a\right)+\left(2011-1\right)\left(2011^{2009}-b\right)\)

A = 2010²⁰¹¹ - 2010²⁰¹⁰

A = 2010²⁰¹⁰ .( 2010 - 1)

A = 2010²⁰¹⁰.2009

2009 ⋮ 2009 ⇒ A = 2010²⁰¹⁰.2009 ⋮ 2009

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)

\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)

\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)