Có : \(\frac{2011}{2012}=\frac{2012-1}{2012}=1-\frac{1}{2012}\)

Có : \(\frac{2012}{2013}=\frac{2013-1}{2013}=1-\frac{1}{2013}\)

Có : \(\frac{2013}{2011}=\frac{2011+2}{2011}=1+\frac{2}{2011}\)

Cộng vế với vế ta có : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{2}{2011}=1+1+1-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)=3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)\)

Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}>0\) nên \(3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)<3\)

Vậy \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}<3\)

\(\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)

Vì \(\frac{2010}{2011+2012+2013}<\frac{2010}{2011};\frac{2011}{2011+2012+2013}<\frac{2011}{2012};\frac{2012}{2011+2012+2013}<\frac{2012}{2013}\)

nên phép dưới nhỏ hơn phép trên

3 lớn nhất

Ta có :

B = \(\dfrac{2011}{2012}\) + \(\dfrac{2012}{2013}\) .

       \(\dfrac{2011}{2012}\) > \(\dfrac{2011}{2012+2013}\) .

        \(\dfrac{2012}{2013}\) > \(\dfrac{2012}{2012+2013}\) .

\(\Rightarrow\) A < B .

Ta có :

B = 2012201320122013 .

       20112012+201320112012+2013 .

        20122012+201320122012+2013 .

 A < B .

Giải:

Ta có:

\(A=\dfrac{2011+2012}{2012+2013}\) 

\(A=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\) 

Vì \(\dfrac{2011}{2012}>\dfrac{2011}{2012+2013}\)  

\(\dfrac{2012}{2013}>\dfrac{2012}{2012+2013}\)

\(\Rightarrow A< B\)

Câu trả lời:A<B

bạn tham khảo:

2010/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2011/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2012/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

=> 2010/2011+2011/2012+2012/2013 > 2010+2011+2012/2011+2012+2013

2010/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2011/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

2012/2011+2012+2013 > 2010+2011+2012/2011+2012+2013

=> 2010/2011+2011/2012+2012/2013 > 2010+2011+2012/2011+2012+2013

\(\overline{ }\)

Bài nãy sai rồi, cho mình làm lại nha:

\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)

\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}\)

Vì: \(\frac{1}{2011}>\frac{1}{2012}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2012}+\frac{1}{2012}>0\)

\(\Rightarrow\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)

Nên \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)

chịu........

Áp dụng tỉ dãy số bằng nhau, ta có:

\(\frac{2011+2012-2013}{2012+2013-2011}=\frac{2011-2012+2013}{2012+2013-2011}=\frac{2011-2012+2013}{-2011-2012+2013}=\left(-1\right)\)

\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}+\dfrac{2013}{2011}\)

=1-\(\dfrac{1}{2011}\)+1\(-\dfrac{1}{2012}\)+1-\(\dfrac{1}{2013}\)+1-\(\dfrac{1}{2011}\)

=4-(\(\dfrac{2}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\)) < 4 

m=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}+\dfrac{2013}{2011}\)

=\(1-\dfrac{1}{2011}+1-\dfrac{1}{2012}+1-\dfrac{1}{2013}+1+\dfrac{2}{2011}\)

=4+\(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\)

vì:

do \(\dfrac{1}{2011}< 1\)

\(\dfrac{1}{2012}< 1\)

\(\dfrac{1}{2013}< 1\)

nên \(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}< 1-1-1=-1\)

hay \(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}< 0\)

nên 4+\(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}< 4\)

vậy tổng m <4

bài này mình tưởng phải lên cấp 2 mới có thế mà mấy em lớp 4 đã phải làm á