Đáp án:

c. \( - \dfrac{3}{2} < m < 2\)

Giải thích các bước giải:

\(\begin{array}{l}
a.Thay:m = 3\\
Hpt \to \left\{ \begin{array}{l}
2x + y = 4\\
2x + 5y = 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
5y - y = 2 - 4\\
2x + y = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  - \dfrac{1}{2}\\
x = \dfrac{9}{4}
\end{array} \right.\\
b.\left\{ \begin{array}{l}
y = 4 - 2x\\
\left( {m - 1} \right)x + \left( {m + 2} \right)\left( {4 - 2x} \right) = 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 4 - 2x\\
\left( {m - 1} \right)x + 4m + 8 + \left( { - 2m - 4} \right)x = 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 4 - 2x\\
\left( {m - 1 - 2m - 4} \right)x =  - 6 - 4m
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 4 - 2x\\
x = \dfrac{{ - 6 - 4m}}{{ - m - 5}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 4 + \dfrac{{12 + 8m}}{{ - m - 5}}\\
x = \dfrac{{ - 6 - 4m}}{{ - m - 5}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{ - 4m - 20 + 12 + 8m}}{{ - m - 5}}\\
x = \dfrac{{6 + 4m}}{{m + 5}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{4m - 8}}{{ - m - 5}}\\
x = \dfrac{{6 + 4m}}{{m + 5}}
\end{array} \right.\left( {DK:m \ne  - 5} \right)\\
Có:S = x + y = \dfrac{{6 + 4m}}{{m + 5}} + \dfrac{{4m - 8}}{{ - m - 5}}\\
 = \dfrac{{6 + 4m - 4m + 8}}{{m + 5}}\\
 = \dfrac{{14}}{{m + 5}}\\
Để:S \in Z\\
 \to \dfrac{{14}}{{m + 5}} \in Z\\
 \to m + 5 \in U\left( {14} \right)\\
 \to \left[ \begin{array}{l}
m + 5 = 14\\
m + 5 =  - 14\\
m + 5 = 7\\
m + 5 =  - 7\\
m + 5 = 2\\
m + 5 =  - 2\\
m + 5 = 1\\
m + 5 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
m = 9\\
m =  - 19\\
m = 2\\
m =  - 12\\
m =  - 3\\
m =  - 7\\
m =  - 4\\
m =  - 6
\end{array} \right.
\end{array}\)

c. Do nghiệm nằm trong góc phần tư thứ nhất

\(\begin{array}{l}
 \to \left\{ \begin{array}{l}
x > 0\\
y > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{4m - 8}}{{ - m - 5}} > 0\\
\dfrac{{6 + 4m}}{{m + 5}} > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{ - 4m + 8}}{{m + 5}} > 0\\
\dfrac{{6 + 4m}}{{m + 5}} > 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m + 5 > 0\\
4m + 6 > 0\\
8 - 4m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 5 < 0\\
4m + 6 < 0\\
8 - 4m < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m >  - 5\\
m >  - \dfrac{3}{2}\\
m < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
m <  - 5\\
m <  - \dfrac{3}{2}\\
m > 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
KL: - \dfrac{3}{2} < m < 2
\end{array}\)