\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)
\(\Leftrightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+5+2012}{2012}+\frac{x+4+2013}{2013}=\frac{x+3+2014}{2014}+\frac{x+2+2015}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)
\(\Rightarrow x+2017=0\)
\(\Rightarrow x=-2017\)
\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\cdot\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\)
Vì \(\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\ne0\)
suy ra \(x+2017=0\)
suy ra \(x=-2017\)
Vậy \(x=-2017\)
Ta có:
\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)
\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+5}{2012}+\frac{2012}{2012}+\frac{x+4}{2013}+\frac{2013}{2013}=\frac{x+3}{2014}+\frac{2014}{2014}+\frac{x+2}{2015}+\frac{2015}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
\(=>\orbr{\begin{cases}x+2017=0\\\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}=0\end{cases}}\)
Mà: \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)
=> x + 2017 = 0 => x = -2017
