Đáp án: $k \in \left\{ { - 2; - 1;1} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
\left( {k - 1} \right).x + y = 3k - 4\\
x + \left( {k - 1} \right)y = k
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {k - 1} \right).x + y = 3k - 4\\
\left( {k - 1} \right)x + {\left( {k - 1} \right)^2}y = k\left( {k - 1} \right)
\end{array} \right.\\
\Rightarrow {\left( {k - 1} \right)^2}.y - y = k\left( {k - 1} \right) - \left( {3k - 4} \right)\\
\Rightarrow \left( {{k^2} - 2k} \right).y = {k^2} - k - 3k + 4\\
\Rightarrow k\left( {k - 2} \right).y = {k^2} - 4k + 4\\
\Rightarrow k\left( {k - 2} \right).y = {\left( {k - 2} \right)^2}\left( * \right)
\end{array}$
Hệ có nghiệm nguyên thì pt (*) có nghiệm nguyên
$\begin{array}{l}
\Rightarrow \left\{ \begin{array}{l}
k \ne 0;k \ne 2\\
y = \dfrac{{k - 2}}{k} \in Z
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
k \ne 0;k \ne 2\\
y = 1 - \dfrac{2}{k} \in Z
\end{array} \right.\\
\Rightarrow \dfrac{2}{k} \in Z\\
\Rightarrow k \in \left\{ { - 2; - 1;1} \right\}\left( {do:k \ne 2} \right)
\end{array}$
