Đáp án:
b) \(\left[ \begin{array}{l}
m = 2\\
m = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x - my = 0\\
mx - y = m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = my\\
m.my - y = m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = my\\
\left( {{m^2} - 1} \right)y = m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = my\\
y = \dfrac{{m + 1}}{{\left( {m - 1} \right)\left( {m + 1} \right)}}
\end{array} \right.\left( {DK:m \ne \pm 1} \right)\\
\to \left\{ \begin{array}{l}
y = \dfrac{1}{{m - 1}}\\
x = \dfrac{m}{{m - 1}}
\end{array} \right.\\
b)x = \dfrac{m}{{m - 1}} = \dfrac{{m - 1 + 1}}{{m - 1}} = 1 + \dfrac{1}{{m - 1}}\\
Do:\left\{ \begin{array}{l}
x \in Z\\
y \in Z
\end{array} \right.\\
\to \dfrac{1}{{m - 1}} \in Z\\
\to m - 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
m - 1 = 1\\
m - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
m = 2\\
m = 0
\end{array} \right.
\end{array}\)
