2. Solve the system of equations: $\left\{\begin{array}{l}x^{2012}+y^{2012}=2, \\ 2012 x-y^{2011}=2012 y-x^{2011}\end{array}\right.$.
2. From the second equation, we get $2012(x-y)=y^{2011}-x^{2011}$. Let's show that $x=y$. Indeed, if, for example, $x>y$, then $y^{2011}>x^{2011}$ and $y>x$ - contradiction. Therefore, $x=y$. From the first equation, we get $2 x^{2012}=2$, from which $x= \pm 1$. Answer: $x=y= \pm 1$.