Đáp án:
a) \(m \in \left( { - \infty ;\,\, - 5} \right) \cup \left( {1; + \infty } \right).\)
b) \(m \in \left( { - \infty ; - 5} \right) \cup \left( {1;\,\,\frac{{21}}{{20}}} \right).\)
Giải thích các bước giải:
Bài 4:
\(\left( {m - 1} \right){x^2} - \left( {2m - 1} \right)x + m + 5 = 0\,\,\,\,\,\,\left( * \right)\)
- a) Phương trình (*) có hai nghiệm trái dấu \( \Leftrightarrow ac < 0\)
\( \Leftrightarrow \left( {m - 1} \right)\left( {m + 5} \right) < 0 \Leftrightarrow \left[ \begin{array}{l}m > 1\\m < - 5\end{array} \right..\)
Vậy \(m \in \left( { - \infty ;\,\, - 5} \right) \cup \left( {1; + \infty } \right).\)
- b) Phương trình (*) có hai nghiệm phân biệt \( \Leftrightarrow \left\{ \begin{array}{l}\Delta > 0\\ - \frac{b}{a} > 0\\\frac{c}{a} > 0\end{array} \right.\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}m - 1 \ne 0\\{\left( {2m - 1} \right)^2} - 4\left( {m - 1} \right)\left( {m + 5} \right) > 0\\\frac{{2m - 1}}{{m - 1}} > 0\\\frac{{m + 5}}{{m - 1}} > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne 1\\4{m^2} - 4m + 1 - 4{m^2} - 16m + 20 > 0\\\left[ \begin{array}{l}m > 1\\m < \frac{1}{2}\end{array} \right.\\\left[ \begin{array}{l}m > 1\\m < - 5\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m \ne 1\\ - 20m + 21 > 0\\\left[ \begin{array}{l}m > 1\\m < - 5\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne 1\\m < \frac{{21}}{{20}}\\\left[ \begin{array}{l}m > 1\\m < - 5\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}1 < m < \frac{{21}}{{20}}\\m < - 5\end{array} \right..\end{array}\)
Vậy \(m \in \left( { - \infty ; - 5} \right) \cup \left( {1;\,\,\frac{{21}}{{20}}} \right).\)
