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12 tháng 7 2017

Đặt \(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}\)\(B=\dfrac{2009^{2007}+1}{2009^{2008}+1}\)

Ta có:

\(2009A=\dfrac{2009.\left(2009^{2008}+1\right)}{2009^{2009}+1}=\dfrac{2009^{2009}+2009}{2009^{2009}+1}\)

\(=\dfrac{2009^{2009}+1+2008}{2009^{2009}+1}=\dfrac{2009^{2009}+1}{2009^{2009}+1}+\dfrac{2008}{2009^{2009}+1}\)

\(=1+\dfrac{1}{2009^{2009}+1}\)

\(2009B=\dfrac{2009.\left(2009^{2007}+1\right)}{2009^{2008}+1}=\dfrac{2009^{2008}+2009}{2009^{2008}+1}\)

\(=\dfrac{2008^{2008}+1+2008}{2009^{2008}+1}=\dfrac{2008^{2008}+1}{2009^{2008}+1}+\dfrac{2008}{2009^{2008}+1}\)

\(=1+\dfrac{2008}{2009^{2008}+1}\)

\(1+\dfrac{2008}{2009^{2009}+1}< 1+\dfrac{2008}{2009^{2008}+1}\)

Nên \(10A< 10B\) \(\Rightarrow A< B\)

Vậy \(\dfrac{2009^{2008}+1}{2009^{2009}+1}< \dfrac{2009^{2007}+1}{2009^{2008}+1}\)

~ Học tốt ~

12 tháng 7 2017

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}< 1\)

\(\Rightarrow A< \dfrac{2009^{2008}+1+2008}{2009^{2009}+1+2008}\Rightarrow A< \dfrac{2009^{2008}+2009}{2009^{2009}+2009}\Rightarrow A< \dfrac{2009\left(2009^{2007}+1\right)}{2009\left(2009^{2008}+1\right)}\Rightarrow A< \dfrac{2009^{2007}+1}{2009^{2008}+1}=B\)\(\Rightarrow A< B\)

21 tháng 9 2023

\(A=\dfrac{2008^{2008}+1}{2008^{2009}+1}\)

\(2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}\)

\(=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}\)

\(=1+\dfrac{2007}{2008^{2009}+1}\)

\(B=\dfrac{2008^{2007}+1}{2008^{2008}+1}\)

\(2008\cdot B=\dfrac{2008^{2008}+2008}{2008^{2008}+1}\)

\(=\dfrac{2008^{2008}+1+2007}{2008^{2008}+1}\)

\(=1+\dfrac{2007}{2008^{2008}+1}\)

Ta có: \(2008^{2009}+1>2008^{2008}+1\)

\(\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}\)

\(\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}\)

\(\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}\)

hay \(A < B\)

#\(Toru\)

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

\(=\dfrac{2008}{2007}-2\cdot2009-\dfrac{2009}{2007}+2009\cdot2\)

=-1/2007

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

Ta có: 

\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)

\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)

\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)

Từ 3 điều trên suy ra : A < B

6 tháng 4 2017

ta có: \(A=\dfrac{2008^{2009}+2}{2008^{2009}-1}=\dfrac{2008^{2009}-1+3}{2008^{2009}-1}=1+\dfrac{3}{2008^{2009}-1}\)

B=\(\dfrac{2008^{2009}}{2008^{2009}-3}=\dfrac{2008^{2009}-3+3}{2008^{2009}-3}=1+\dfrac{3}{2008^{2009}-3}\)

ta thấy: \(1+\dfrac{3}{2008^{2009}-1}\)<\(1+\dfrac{3}{2008^{2009}-3}\)

vậy A<B