Vì 20009 x 2009 + 2008 < 2009 x 2009 + 2009

=>A < 1

Ta có: \(B=\frac{2009x2009+2009}{2008x2009+2010}=\frac{2009x\left(2008+1\right)+2009}{2008x2009+2010}=\frac{2008x2009+2009+2009}{2008x2009+2010}\)

\(B=\frac{2008x2009+4018}{2008x2009+2010}=\frac{2008x2009+2010+2008}{2008x2009+2010}=\frac{2008x2009+2010}{2008x2009+2010}+\frac{2008}{2008x2009+2010}\)

\(B=1+\frac{2008}{2008x2009+2010}>1\)

Mà A < 1

=>A < B

dòng đầu sửa chỗ 20009 nhé,2009

- Uầy:)) Mới đây 3 li-ke:)) Nhanh vcc

mình cũng có bài giống như này nhưng chưa làm được

mình cũng thế

^{Ta có: 2008/2009 > 2008/2009+2010+2011}

       2009/2010> 2009/2010+2011

      2010/2011>2010>2010/2009+2010+2011

Suy ra: A>2008+2009+2010/2009+2010+2011

Vậy A >B

Tớ cũng có bài này nhưng chưa làm được

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Ta có :

�=20092010−220092011−2<1B=20092011−220092010−2​<1

⇔�<20092010−2+201120092011−2+2011=20092010+200920092011+2009=2009(20092009+1)2009(20092010+1)=20092009+120092010+1=�⇔B<20092011−2+201120092010−2+2011​=20092011+200920092010+2009​=2009(20092010+1)2009(20092009+1)​=20092010+120092009+1​=A

$\iff A>B$

Ta có :

\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)

\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)

\(\Leftrightarrow A>B\)

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