\(\frac{2009}{2010}\) +\(\frac{2010}{2011}\)...">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

9 tháng 6 2020

Ta có: \(\frac{2009}{2010}>\frac{2009}{2010+2011}\)  ; \(\frac{2010}{2011}>\frac{2010}{2011+2010}\)

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

=> A > B

9 tháng 6 2020

Ta có  \(\frac{2009}{2010}>\frac{2009}{2010+2011}\)           ,   \(\frac{2010}{2011}>\frac{2010}{2011+2010}\)

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

\(\Rightarrow A>B\)

10 tháng 6 2020

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

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

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

10 tháng 6 2020

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

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

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

16 tháng 1 2019

A=2.998508205

B=0.999502735

suy ra A>B

30 tháng 5 2019

                                              Bài giải

Theo bài ra :  

\(A=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}\)

\(B=\frac{2009+2010+2011}{2010+2011+2012}=\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)

Ta có : 

\(\frac{2009}{2010}>\frac{2009}{2010+2011+2012}\)

\(\frac{2010}{2011}>\frac{2010}{2010+2011+2012}\)

\(\frac{2011}{2012}>\frac{2011}{2010+2011+2012}\)

\(\Rightarrow\text{ }\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}>\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)

\(\Rightarrow\text{ }A>B\)

9 tháng 3 2018

\(b)\)  Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)

Vậy \(\frac{2009^{2009}+1}{2009^{2010}+1}>\frac{2009^{1010}-2}{2009^{2011}-2}\)

Chúc bạn học tốt ~

9 tháng 3 2018

Àk mình còn thiếu một điều kiện nữa xin lỗi nhé : 

Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Bạn thêm vào nhé 

12 tháng 1 2019

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

\(\Rightarrow B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}\)\(=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)

Suy ra : \(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2009}+1}{2009^{2010}+1}\) hay \(B< A\)

Vậy \(A>B\)

25 tháng 5 2018

Ta có :

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{20101}{2009+2010+2011}\)

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

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

Suy ra : A > B

7 tháng 7 2017

Do 2009\(^{2010}\)-2 < 2009\(^{2011}\)-2 \(\Rightarrow\)B<1

Theo đề bài ta có: 

B= \(\frac{2009^{2010}-2}{2009^{2011}-2}\)\(\frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)\(\frac{2009.\left(1+2009^{2009}\right)}{2009.\left(1+2009^{2010}\right)}\)\(\frac{2009^{2009}+1}{2009^{2010}+1}\)= A \(\Rightarrow\)B<A