Đáp án:

\[\left[ \begin{array}{l}
x = 2;\,\,y = 3\\
x = 3;\,\,y = 2
\end{array} \right.\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {y - 1} \right)\left( {x + y - 2} \right) = 6\\
{x^2} + {y^2} - 2x - 2y - 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {y - 1} \right)\left( {\left( {x - 1} \right) + \left( {y - 1} \right)} \right) = 6\\
\left( {{x^2} - 2x + 1} \right) + \left( {{y^2} - 2y + 1} \right) - 5 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {y - 1} \right)\left( {\left( {x - 1} \right) + \left( {y - 1} \right)} \right) = 6\\
{\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 5
\end{array} \right.
\end{array}\)

Đặt \(a = x - 1;\,\,\,b = y - 1\), khi đó, hệ phương trình trên trở thành:

\(\begin{array}{l}
\left\{ \begin{array}{l}
ab.\left( {a + b} \right) = 6\\
{a^2} + {b^2} = 5
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^2} - 2ab = 5
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {a,b \ne 0} \right)\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^2} - 2.\dfrac{6}{{a + b}} = 5
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^3} - 12 = 5\left( {a + b} \right)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^3} - 5\left( {a + b} \right) - 12 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
\left[ {{{\left( {a + b} \right)}^3} - 3.{{\left( {a + b} \right)}^2}} \right] + \left[ {3.{{\left( {a + b} \right)}^2} - 9\left( {a + b} \right)} \right] + \left[ {4.\left( {a + b} \right) - 12} \right] = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
\left[ {\left( {a + b} \right) - 3} \right]\left[ {{{\left( {a + b} \right)}^2} + 3.\left( {a + b} \right) + 4} \right] = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
a + b = 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a + b = 3\\
ab = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 3 - a\\
a.\left( {3 - a} \right) = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 3 - a\\
{a^2} - 3a + 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 3 - a\\
\left( {a - 1} \right)\left( {a - 2} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 1;\,\,b = 2\\
a = 2;\,\,b = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2;\,\,y = 3\\
x = 3;\,\,y = 2
\end{array} \right.
\end{array}\)