Có bao nhiêu cặp số nguyên \(\left( {x;y} \right)\) thỏa mãn  \(xy = 2\left( {x + y} \right)\).

Ta có \(xy = 2\left( {x + y} \right) \Leftrightarrow 2x + 2y - xy = 0\) \( \Leftrightarrow 2x - xy + 2y - 4 =  - 4\)\( \Leftrightarrow x\left( {2 - y} \right) + 2\left( {y - 2} \right) =  - 4\)

\( \Leftrightarrow \left( {x + 2} \right)\left( {2 - y} \right) =  - 4 \Leftrightarrow \left( {x + 2} \right)\left( {y - 2} \right) = 4\)

Mà \(x;y \in \mathbb{Z} \Rightarrow \left( {x + 2} \right);\left( {y - 2} \right) \in Ư\left( 4 \right) \)\(= \left\{ { - 1;1; - 2;2; - 4;4} \right\}\)

+ \(\left\{ \begin{array}{l}x + 2 =  - 1\\y - 2 =  - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x =  - 3\\y =  - 2\end{array} \right.\)

+ \(\left\{ \begin{array}{l}x + 2 = 1\\y - 2 = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x =  - 1\\y = 6\end{array} \right.\)

+ \(\left\{ \begin{array}{l}x + 2 = 2\\y - 2 = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 0\\y = 4\end{array} \right.\)

+ \(\left\{ \begin{array}{l}x + 2 =  - 2\\y - 2 =  - 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x =  - 4\\y = 0\end{array} \right.\)

+ \(\left\{ \begin{array}{l}x + 2 =  - 4\\y - 2 =  - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x =  - 6\\y = 1\end{array} \right.\)

+ \(\left\{ \begin{array}{l}x + 2 = 4\\y - 2 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = 3\end{array} \right.\)

Vậy có 6 cặp số \(\left( {x;y} \right)\) thỏa mãn điều kiện đề bài.