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Ta có :
\(\dfrac{n}{n+1}=1-\dfrac{1}{n+1}\)
\(\dfrac{n+2012}{n+2013}=1-\dfrac{1}{n+2013}\)
Ta có : \(\dfrac{1}{n+1}>\dfrac{1}{n+2013}\) (vì 1>0; 0<n+1<n+2013
=> \(-\dfrac{1}{n+1}< -\dfrac{1}{n+2013}\)
=> \(1-\dfrac{1}{n+1}< 1-\dfrac{1}{n+2013}\)
=> \(\dfrac{n}{n+1}< \dfrac{n+2012}{n+2013}\)
Vậy ...
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta thấy: \(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\)
\(\Rightarrow M=\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}>N=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Vậy M>N
\(\dfrac{n}{n+1}=\dfrac{\left(n+1\right)-1}{n+1}=1-\dfrac{1}{n+1}\)
\(\dfrac{n+2013}{n+2012}=\dfrac{\left(n+2013\right)-1}{n+2013}=1-\dfrac{1}{n+2013}\)
Ta có:
\(n+1< n+2013\Rightarrow\dfrac{1}{n+1}>\dfrac{1}{n+2013}\Rightarrow\dfrac{-1}{n+1}\)
\(< \dfrac{-1}{n+2013}\)
\(\Rightarrow1-\dfrac{1}{n+1}< 1-\dfrac{1}{n+2013}\) hay \(\dfrac{n}{n+1}< \dfrac{n+2012}{n+2013}\)