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27 tháng 1 2019

1.

ta có: 2009A= (2009^2010+ 2009)/ (2009^2010+1)= (2009^10+1+2008)/(2009^2010+1)=1+ [2008/(2009^2010+1)]

làm tương tự như trên ta được :

2009B=1-[4016/(2009^2011-2)]

lại có:

2009A= .............(nt) > 1

2009B=...........<1

=>2009A>2009B

=>A>B

27 tháng 1 2019

câu 2 và 3 thì làm sao bạn

27 tháng 1 2021

Ta có :

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

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

\(\Leftrightarrow A>B\)

26 tháng 2 2018

Ta có : 

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

Vì : 

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

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

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

Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

26 tháng 2 2018

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

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

Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)

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

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

nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)

hay A > B

Vậy A > B 

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

1 tháng 4 2021

A=-2015/2015x2016

A=-1/2016

B=-2014/2014x2015

B=-1/2015

vi 2016>2015,-1/2016>-1/2015

vay A>B

b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)

\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)

Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)

\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)

Ta có: \(10^{2010}+1< 10^{2011}+1\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)

\(\Leftrightarrow10A>10B\)

hay A>B

26 tháng 2 2018

\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)

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

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

Hay A > B

23 tháng 12 2018

bằng nhau bạn nhé

13 tháng 2 2019

=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)

\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)

15 tháng 3 2017

A=\(\dfrac{2009^{2010}+1}{2009^{2009}+1}\)

2009A=\(\dfrac{(2009^{2010}+1)+0}{2009^{2010}+1}\)

= 1+\(\dfrac{0}{2009^{2010}+1}\)= 1+0 =1

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

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

=\(\dfrac{(2009^{2011}-1)-0}{2009^{2011}-2009}\)

= \(1-\dfrac{0}{2009^{2011}-2009}\)

=1-0= 1

Vì 1=1\(\Rightarrow A=B\)

16 tháng 4 2017

Ta có : A = 2009^2010+1/2009^2009+1

Suy ra: 1/2009 A = 1 - 2008/2009^2010+2009 (1)

Lại có:B = 2009^2011 - 2 / 2009^2010 - 2

Suy ra : 1/2009 B = 1 + 4016/2009^2011-4018 (2)

Vì 1 - 2008/2009^2010+2009 < 1 + 4016/2009^2011-4018 (3)

Từ (1);(2) và (3) suy ra : 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