Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\left( {{d_1}} \right):y = mx - 2\\
\left( {{d_2}} \right):y = \frac{9}{m}x - \frac{3}{m} - 1\\
a.m = - 2\\
\to \left\{ \begin{array}{l}
- 2x - y = 2\\
9x + 2y = 1
\end{array} \right. \to \left\{ \begin{array}{l}
x = 1\\
y = - 4
\end{array} \right.\\
b.\left( {{d_1}} \right)//\left( {{d_2}} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{9}{m} = m\left( {m \ne 0} \right)\\
- \frac{3}{m} - 1 \ne - 2
\end{array} \right. \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m = 3\left( l \right)\\
m = - 3
\end{array} \right.\\
m \ne 3
\end{array} \right.\\
\to m = - 3\left( {TM} \right)\\
c.\left( {{d_1}} \right) \equiv \left( {{d_2}} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{9}{m} = m\left( {m \ne 0} \right)\\
- \frac{3}{m} - 1 = - 2
\end{array} \right. \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m = 3\\
m = - 3
\end{array} \right.\\
m = 3
\end{array} \right.\\
\to m = 3\left( {TM} \right)\\
d. \Leftrightarrow \left\{ \begin{array}{l}
\frac{9}{m} \ne m\\
x > 0\\
y > 0
\end{array} \right.\\
Có:\left\{ \begin{array}{l}
m \ne \pm 3\\
mx - 2 = \frac{9}{m}x - \frac{3}{m} - 1\\
y = mx - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \pm 3\\
\left( {m - \frac{9}{m}} \right)x = 1 - \frac{3}{m}\left( * \right)\\
y = mx - 2
\end{array} \right.\\
\left( * \right) \to x = \frac{{m - 3}}{m}:\frac{{\left( {m - 3} \right)\left( {m + 3} \right)}}{m} = \frac{1}{{m + 3}}\\
\to y = \frac{{m - 2m - 6}}{{m + 3}} = \frac{{ - m - 6}}{{m + 3}}\\
\to \left\{ \begin{array}{l}
m \ne \pm 3\\
\frac{1}{{m + 3}} > 0\\
\frac{{ - m - 6}}{{m + 3}} > 0
\end{array} \right. \to \left\{ \begin{array}{l}
m \ne \pm 3\\
m + 3 > 0\\
- m - 6 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \pm 3\\
m > - 3\\
m < - 6
\end{array} \right.\\
\to m \in \emptyset \\
e. \Leftrightarrow \left\{ \begin{array}{l}
m \ne \pm 3\\
x < 0\\
y > 0
\end{array} \right. \to \left\{ \begin{array}{l}
m \ne \pm 3\\
\frac{1}{{m + 3}} < 0\\
\frac{{ - m - 6}}{{m + 3}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \pm 3\\
m + 3 < 0\\
- m - 6 < 0
\end{array} \right. \to \left\{ \begin{array}{l}
m \ne \pm 3\\
m < - 3\\
m > - 6
\end{array} \right.\\
\to - 6 < m < - 3
\end{array}\)
