By factorising

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

$$\begin{align}&(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{align}$$

Expanding both brackets and collecting like terms gives

$$\begin{align}&(x-2013)(x-2012)-(x-2012)(x-2011)=0\\

\Rightarrow&(x^2-2013x-2012x+2012\times2013)-(x^2-2012x-2011x+2011\times2012)=0\\

\Rightarrow&x^2-2013x-2012x+2012\times2013-x^2+2012x+2011x-2011\times2012=0\\

\Rightarrow&-2x+2012\times2013-2012\times2011=0\\

\Rightarrow&-2x+2\times2012=0\\

\Rightarrow&2\times2012=2x\\

\Rightarrow&2012=x\end{align}$$

You can find more short problems, arranged by curriculum topic, in our short problems collection.