Đáp án:
a) $\left\{\begin{array}{l}
m = 31\\
m = 7
\end{array} \right.$
b) $m=\{7,-17,1,-11,-3,-7,-4,-6\}$
Giải thích các bước giải:
\(\begin{array}{l}
a.\left\{ \begin{array}{l}
x +y=1\\
3x+(m-1)y= 12\\
(m-1)x+12y = 24
\end{array} \right. \to \left\{ \begin{array}{l}
x = 1 - y\\
3 - 3y + my - y = 12\\
m - my + y - 1 + 12y = 24
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - y\\
y = \frac{9}{{m - 4}}\\
m - \frac{{9m}}{{m - 4}} + \frac{9}{{m - 4}} + \frac{{108}}{{m - 4}} = 25\left( * \right)
\end{array} \right.\\
\left( * \right) \to {m^2} - 4m - 9m + 117 = 25m - 100\\
\to \left[ \begin{array}{l}
m = 31\\
m = 7
\end{array} \right.
\end{array}\)
b. Để hệ có nghiệm duy nhất là nghiệm nguyên
\(\begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{{12 - \left( {m - 1} \right)y}}{3}\\
\frac{{12 - \left( {m - 1} \right)y}}{3}.\left( {m - 1} \right) + 12y = 24\left( * \right)
\end{array} \right.\\
\left( * \right) \to 12m - 12 - {\left( {m - 1} \right)^2}y + 36y = 72\\
\to y\left[ {36 - {{\left( {m - 1} \right)}^2}} \right] = 84 - 12m\\
\to y\left( {7 - m} \right)\left( {5 + m} \right) = 12\left( {7 - m} \right)\\
\to y = \frac{{12}}{{5 + m}}\\
\to x = \frac{{12 - \left( {m - 1} \right)y}}{3} = \frac{{12 - \frac{{12m - 12}}{{5 + m}}}}{3} = \frac{{60 + 12m - 12m + 12}}{{15 + 3m}} = 2.\frac{{12}}{{5 + m}}\\
Do:\left\{ \begin{array}{l}
x \in Z\\
y \in Z
\end{array} \right. \to \left[ \begin{array}{l}
5 + m = 12\left( l \right)\\
5 + m = - 12\left( l \right)\\
5 + m = 6\left( l \right)\\
5 + m = - 6\left( l \right)\\
5 + m = 2\\
5 + m = - 2\left( l \right)\\
5 + m = 1\left( l \right)\\
5 + m = - 1
\end{array} \right. \to \left[ \begin{array}{l}
m=7\\m=-17\\m=1\\m=-11\\m=-3\\m=-7\\m=-4\\m=-6
\end{array} \right.
\end{array}\)
