Bài giải :

${{\left| x-3 \right|}^{2014}}+{{\left| 6+2y \right|}^{2015}}\le 0$

Nhận xét :

${{\left| x-3 \right|}^{2014}}\ge 0$

${{\left| 6+2y \right|}^{2015}}\ge 0$

Để ${{\left| x-3 \right|}^{2014}}+{{\left| 6+2y \right|}^{2015}}\le 0$  thì  $\left\{ \begin{align}& {{\left| x-3 \right|}^{2014}}=0 \\ & {{\left| 6+2y \right|}^{2015}}=0 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x-3=0 \\ & 6+2y=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=3 \\  & y=-3 \\ \end{align} \right.$

Suy ra (x; y) = (3; -3)

Vậy x+y=0.