$|x-2006y| + |x-2012| ≤ 0$
Vì : $|x-2006y|;|x-2012| ≥ 0 ∀ x;y$
$⇒$ $\left\{\begin{matrix} x - 2006y = 0& \\ x - 2012 = 0& \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix} y = \dfrac{1006}{1003}& \\ x = 2012 & \end{matrix}\right.$
Vậy `(x;y)=(2012;{1006}/{1003})`