Đặt \(\left\{ \begin{array}{l}x - 1 = u\\y + 2 = v\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = u + 1\\y = v - 2\end{array} \right.\), hệ trở thành

\(\left\{ \begin{array}{l}{u^2} + {v^2} = 9\\\left( {u + 1} \right)\left( {v - 2} \right)\left( {u - 1} \right)\left( {v + 2} \right) =  - 5\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{u^2} + {v^2} = 9\\\left( {{u^2} - 1} \right)\left( {{v^2} - 4} \right) =  - 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {{u^2} - 1} \right) + \left( {{v^2} - 4} \right) = 4\\\left( {{u^2} - 1} \right)\left( {{v^2} - 4} \right) =  - 5\end{array} \right.\)

Đặt \(\left\{ \begin{array}{l}a = {u^2} - 1\\b = {v^2} - 4\end{array} \right.\) ta được \(\left\{ \begin{array}{l}a + b = 4\\ab =  - 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 5\\b =  - 1\end{array} \right.\\\left\{ \begin{array}{l}a =  - 1\\b = 5\end{array} \right.\end{array} \right.\)

Suy ra \(\left\{ \begin{array}{l}{u^2} - 1 = 5\\{v^2} - 4 =  - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u^2} = 6\\{v^2} = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}u =  \pm \sqrt 6 \\v =  \pm \sqrt 3 \end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = 1 \pm \sqrt 6 \\y =  - 2 \pm \sqrt 3 \end{array} \right.\)

Hoặc \(\left\{ \begin{array}{l}{u^2} - 1 =  - 1\\{v^2} - 4 = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u^2} = 0\\{v^2} = 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}u = 0\\v =  \pm 3\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = 1\\y = 1,y =  - 5\end{array} \right.\)