\(\text{Đặt : }A=\frac{2009^{2008}+1}{2009^{2009}+1}\Rightarrow2009A=\frac{2009^{2009}+2009}{2009^{2009}+1}=1+\frac{2008}{2009^{2009}+1}\)

\(B=\frac{2009^{2007}+1}{2009^{2008}+1}\Rightarrow2009B=\frac{2009^{2008}+2009}{2009^{2008}+1}=1+\frac{2008}{2009^{2008}+1}\)

Ta thấy: \(\frac{2008}{2009^{2009}+1}<\frac{2008}{2009^{2008}+1}\)

=>2009A<2009B =>A<B

Hay \(\frac{2009^{2008}+1}{2009^{2009}+1}<\frac{2009^{2007}+1}{2009^{2008}+1}\)

b)Đề chưa rõ xem lại

Đặt : A=20092008+120092009+1⇒2009A=20092009+200920092009+1=1+200820092009+1Đặt : A=20092008+120092009+1⇒2009A=20092009+200920092009+1=1+200820092009+1

B=20092007+120092008+1⇒2009B=20092008+200920092008+1=1+200820092008+1B=20092007+120092008+1⇒2009B=20092008+200920092008+1=1+200820092008+1

Ta thấy: 200820092009+1<200820092008+1200820092009+1<200820092008+1

=>2009A<2009B =>A<B

Hay 20092008+120092009+1<20092007+120092008+1

Giải:

Ta có:

A=2009²⁰⁰⁸+1/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+2009/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+1+2008/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+1/2009²⁰⁰⁹+1 + 2008/2009²⁰⁰⁹+1

2009A=1+2008/2009²⁰⁰⁹+1

Tương tự:

B=2009²⁰⁰⁹+1/2009²⁰¹⁰+1

2009B=1+2008/2009²⁰¹⁰+1

Vì 2008/2009²⁰⁰⁹+1 > 2008/2009²⁰¹⁰+1 nên 2009A>2009B

⇒A>B

2009²⁰⁰⁸ +1/2009²⁰⁰⁹ +1 >....

de qua viet moi tay

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

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

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

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

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

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

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

ta có A = 2008^2009+2 / 2008^2009-1 = 2008^2009-1+3 / 2008^2009-1 = 1 + 3/2008^2009-1

lại có B = 2008^2009 / 2008^2009-3 = 2008^2009-3+3 / 2008^2009-3 = 1 + 3/2008^2009-3

vì 3/2008^2009-1 < 3/2008^2009-3 => 1 + 3/2008^2009-1 < 1 + 3/2008^2009-3

Hay A<B 

Vậy A<B

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