$\begin{array}{l} A = 2{x^4}{y^9},B = - \dfrac{1}{2}{x^3}{y^9}\\ A + B = 0,5 = \dfrac{1}{2}\\ \Rightarrow 2{x^4}{y^9} - \dfrac{1}{2}{x^3}{y^9} = \dfrac{1}{2}\\ \Leftrightarrow {x^3}{y^9}\left( {2x - \dfrac{1}{2}} \right) = \dfrac{1}{2}\\ \Leftrightarrow {x^3}{y^9}\left( {4x - 1} \right) = 1\\ \Rightarrow \left\{ \begin{array}{l} {x^3}{y^9} = 1\\ 4x - 1 = 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = \dfrac{1}{2}\\ {y^9} = 2 \end{array} \right.(L)\\ \Rightarrow \left\{ \begin{array}{l} {x^3}{y^9} = - 1\\ 4x - 1 = - 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {x^3}{y^9} = - 1\\ x = 0 \end{array} \right.(L)\\ \Rightarrow \left( {x;y} \right) = \emptyset \end{array}$
Vậy không có cặp số nguyên $(x;y)$ thỏa mãn đề bài
