Thousands of x

Answer: $2012$

By factorising

Notice that the two terms have a common factor of $x-2012$, so

$$\begin{array}{rl}& (x-2013)(x-2012)-(x-2012)(x-2011)=0\\ \Rightarrow & (x-2012)((x-2013)-(x-2011))=0\\ \Rightarrow & (x-2012)(x-2013-x+2011)=0\\ \Rightarrow & (x-2012)\times -2=0\\ \Rightarrow & x-2012=0\\ \Rightarrow & x=2012\end{array}$$

By expanding

Expanding both brackets and collecting like terms gives

$$\begin{array}{rl}& (x-2013)(x-2012)-(x-2012)(x-2011)=0\\ \Rightarrow & (x^2-2013x-2012x+2012\times 2013)-(x^2-2012x-2011x+2011\times 2012)=0\\ \Rightarrow & x^2-2013x-2012x+2012\times 2013-x^2+2012x+2011x-2011\times 2012=0\\ \Rightarrow & -2x+2012\times 2013-2012\times 2011=0\\ \Rightarrow & -2x+2\times 2012=0\\ \Rightarrow & 2\times 2012=2x\\ \Rightarrow & 2012=x\end{array}$$