PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
Ta có: \(\frac{x}{2017}+\frac{x+1}{2016}=\frac{x+2}{2015}+\frac{x+3}{2014}\)
\(\Leftrightarrow\frac{x}{2017}+1+\frac{x+1}{2016}+1=\frac{x+2}{2015}+1+\frac{x+3}{2014}+1\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}-\frac{x+2017}{2015}-\frac{x+2017}{2014}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\)
nên x+2017=0
hay x=-2017
Vậy: S={-2017}
