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

   = 3 + \(\frac{2}{2011}-\frac{1}{2012}-\frac{1}{2013}\)

  có \(\frac{1}{2011}>\frac{1}{2012}\)và \(\frac{1}{2011}>\frac{1}{2013}\)

\(\Rightarrow S>3\)

mai mink phải nộp rồi

may quá! Thanks bạn rất nhiều

S lớn hơn 3 vì , S = 3,000000741

Bài giải : 

Theo đề bài ra ta có : n. (n - 1) : 2 = 435 

=> n. (n - 1) = 435 . 2 = 870 

=> n.(n-1) = 30. 29

Vậy n = 30. 

N la : 30

30

30

30

30

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

Cho s= 2010/2011+2011/2012+2012/2013/2013/2010

Cho s=1

AM-GM:\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2010}\ge4\sqrt[4]{\frac{2010.2011.2012.2013}{2011.2012.2013.2010}}=4\sqrt[4]{1}=4\)

\(\Rightarrow S\ge4\)

^^

\(S=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2010}\ge4^4\sqrt[]{\frac{2010.2011.2012.2013}{2011.2012.2013.2010}=4}\)\(4^4\sqrt[]{1}=4\)

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:

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

Ta có:

Q=2010/2011+2012+2013+2011/2011+2012+2013+2012/2011+2012+2013

Mà 2010/2011+2012+2013<2010/2011

      2011/2011+2012+2013<2011/2012

      2012/2011+2012+2013<2012/2013

=>Q<P

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

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

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

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

=> P > Q

P > Q  không phải toán lớp 6

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

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

Vì: \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)

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

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

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

                    P                         >                                         Q