Để phương trình có hai nghiệm phân biệt trái dấu
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\x_1x_2< 0\end{matrix}\right.\) (1)
Mà theo hệ thức Vi-ét ta có: \(x_1x_2=\frac{c}{a}=-\frac{m}{m+1}\)
Vậy từ (1) và (2) ta có:
\(\Leftrightarrow\left\{{}\begin{matrix}\left(-3m\right)^2-4.\left(m+1\right)\left(-m\right)>0\\-\frac{m}{m+1}< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9m^2+4\left(m^2+m\right)>0\\\left[{}\begin{matrix}\left\{{}\begin{matrix}-m< 0\\m+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}-m>0\\m+1< 0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13m^2+4m>0\\\left[{}\begin{matrix}\left\{{}\begin{matrix}m>0\\m>-1\end{matrix}\right.\\\left\{{}\begin{matrix}m< 0\\m< -1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\left(13m+4\right)>0\\\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}m>0\\13m+4>0\end{matrix}\right.\\\left\{{}\begin{matrix}m< 0\\13m+4< 0\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}m>0\\m>-\frac{4}{13}\end{matrix}\right.\\\left\{{}\begin{matrix}m< 0\\m< -\frac{4}{13}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>0\\m< -\frac{4}{13}\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\end{matrix}\right.\)
Ghép hợp lí ta được \(m>0\) hoặc \(m< -1\)
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