[tex] \sum\limits_{k=1}^{2012} \dfrac{x+k}{k+1} = 2012 \\ \\ \sum\limits_{k=1}^{2012} \dfrac{(x-1)+(k+1)}{k+1} = 2012 \\ \\ \sum\limits_{k=1}^{2012} \Big(\dfrac{x-1}{k+1} +1 \Big) = 2012 \\ \\ \sum\limits_{k=1}^{2012} \Big(\dfrac{x-1}{k+1} \Big)+ 2012 = 2012 \\ \\ (x-1)\cdot \sum\limits_{k=1}^{2012} \dfrac{1}{k+1} = 2012-2012 \\ \\ (x-1)\cdot \sum\limits_{k=1}^{2012} \dfrac{1}{k+1} = 0 \\ \\ x-1 = \dfrac{0}{\sum\limits_{k=1}^{2012} \dfrac{1}{k+1}} \\ \\ x-1 = 0 \Rightarrow \boxed{x = 1}[/tex]
