Bạn tham khảo:
$\dfrac{x+1}{2012} + \dfrac{x+2}{2011} + \dfrac{x+3}{2010} = \dfrac{x-1}{2014} + \dfrac{x-2}{2015} + \dfrac{x-3}{2016$
$⇔ \dfrac{x+1}{2012} + \dfrac{x+2}{2011} + \dfrac{x+3}{2010} + 3 = \dfrac{x-1}{2014} + \dfrac{x-2}{2015} + \dfrac{x-3}{2016} + 3$
$⇔ \dfrac{x+1}{2012} + \dfrac{x+2}{2011} + \dfrac{x+3}{2010} + 1 + 1 + 1 = \dfrac{x-1}{2014} + \dfrac{x-2}{2015} + \dfrac{x-3}{2016} + 1 + 1 + 1$
$⇔ (\dfrac{x+1}{2012}+1) + (\dfrac{x+2}{2011}+1) + (\dfrac{x+3}{2010}+1) = (\dfrac{x-1}{2014}+1) + (\dfrac{x-2}{2015}+1) + (\dfrac{x-3}{2016}+1)$
$⇔ \dfrac{x+2013}{2012} + \dfrac{x+2013}{2011} + \dfrac{x+2013}{2010} = \dfrac{x+2013){2014} + \dfrac{x+2013}{2015} + \dfrac{x+2013}{2016}$
$⇔ \dfrac{x+2013}{2012} + \dfrac{x+2013}{2011} + \dfrac{x+2013}{2010} - \dfrac{x+2013){2014} - \dfrac{x+2013}{2015} - \dfrac{x+2013}{2016} = 0$
$⇔ ( x + 2013 ) ( \dfrac{1}{2012} \dfrac{1}{2011} + \dfrac{1}{2010} - \dfrac{1}{2014} - \dfrac{1}{2015} - \dfrac{1}{2016} )$
Vì $\dfrac{1}{2012} \dfrac{1}{2011} + \dfrac{1}{2010} - \dfrac{1}{2014} - \dfrac{1}{2015} - \dfrac{1}{2016} \neq 0$
$⇔ x + 2013 = 0$
$⇔ x = -2013$
Kết luận tập nghiệm của phương trình trên là $S =$ { $-2013$ }
