Đáp án:
\(\left[ \begin{array}{l}
3 > m > 1\\
m < 1
\end{array} \right.\)
Giải thích các bước giải:
Phương trình hoành độ giao điểm của 2 đường thẳng
\(\begin{array}{l}
\left( {3m - 1} \right)x + 2 - m = 8x - m + 5\\
\to \left( {3m - 9} \right)x = 3\\
\to x = \frac{3}{{3\left( {m - 3} \right)}}\left( {m \ne 3} \right)\\
\to x = \frac{1}{{m - 3}}\\
\to y = 8x - m + 5\\
= \frac{{8 - {m^2} + 3m + 5m - 15}}{{m - 3}}\\
= \frac{{ - {m^2} + 8m - 7}}{{m - 3}}
\end{array}\)
Do 2 đường thẳng cắt nhau tại 1 điểm bên trái trục tung
TH1: x<0;y>0
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
\frac{1}{{m - 3}} < 0\\
\frac{{ - {m^2} + 8m - 7}}{{m - 3}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m - 3 < 0\\
\left( {m - 7} \right)\left( {1 - m} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 3\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 7 > 0\\
1 - m < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 7 < 0\\
1 - m > 0
\end{array} \right.
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
m < 3\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 7\\
m > 1
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 7\\
m < 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
⇒ m < 1
\end{array}\)
TH2: x<0; y<0
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
\frac{1}{{m - 3}} < 0\\
\frac{{ - {m^2} + 8m - 7}}{{m - 3}} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m - 3 < 0\\
\left( {m - 7} \right)\left( {1 - m} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 3\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 7 > 0\\
1 - m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 7 < 0\\
1 - m < 0
\end{array} \right.
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
m < 3\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 7\\
m < 1
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
m < 7\\
m > 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to 3 > m > 1
\end{array}\)
\(KL:\left[ \begin{array}{l}
3 > m > 1\\
m < 1
\end{array} \right.\)
