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Ta có : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)
Mà : \(\left(\frac{2014}{2015}+\frac{1}{2014}\right)>1;\left(\frac{2015}{2016}+\frac{1}{2014}\right)>1;\frac{2014}{2014}=1\)
Nên : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)\(>1+1+1=3\)
Ta có:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)
Mà:\(\left(\frac{2014}{2015}+\frac{1}{2014}\right)>1:\left(\frac{2015}{2016}+\frac{1}{2014}\right)>\)\(1:\frac{2014}{2014}=1\)
Nên:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}>1+1+1=3\)
\(N=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(N=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(N=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(N=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}=\frac{1}{2017}\)
a)
Ta có: \(\frac{x+y}{2014}\ne\frac{x-y}{2016}\)
\(\Leftrightarrow2016x+2016y=2014x-2014y\)
\(\Leftrightarrow2x=-4030y\)
\(\Leftrightarrow x=-2015y\)
Thay \(x=-2015y\)vào \(\frac{x+y}{2014}=\frac{xy}{2015}\)ta được:
\(\Leftrightarrow\frac{-2015+y}{2014}=\frac{-2015y}{2015}\)
\(\Leftrightarrow\frac{-2014y}{2014}=\frac{-2015y^2}{2015}\)
\(\Leftrightarrow-y=-y^2\)
\(\Leftrightarrow y-y^2=0\)
\(\Leftrightarrow y\left(1-y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y=0\\1-y=0\end{cases}}\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)
Trường hợp \(y=0\):
\(y=0\Rightarrow x.y=-2015.0=0\)
Trường hợp \(y=1\):
\(y=1\Rightarrow x.y=-2015.1=-2015\)