\(a,\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\\dfrac{c}{a}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-12\left(2m+1\right)\ge0\\\dfrac{2m+1}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{13}{24}\\m< -\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow m< -\dfrac{1}{2}\)
\(b,\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\dfrac{c}{a}>0\\-\dfrac{b}{a}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{13}{24}\\m>-\dfrac{1}{2}\\-\dfrac{5}{3}< 0\left(đúng\right)\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< m< \dfrac{13}{24}\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\dfrac{c}{a}>0\\-\dfrac{b}{a}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{13}{24}\\m>-\dfrac{1}{2}\\-\dfrac{5}{3}>0\left(\text{vô lí}\right)\end{matrix}\right.\Leftrightarrow m\in\varnothing\)
\(d,\Leftrightarrow\Delta\ge0\Leftrightarrow m\le\dfrac{13}{24}\\ \text{Viét: }\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{3}\\x_1x_2=\dfrac{2m+1}{3}\end{matrix}\right.\\ x_1^3+x_2^3+10\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=10\\ \Leftrightarrow-\dfrac{125}{27}-\left(2m+1\right)\left(-\dfrac{5}{3}\right)=10\\ \Leftrightarrow\dfrac{5}{3}\left(2m+1\right)=\dfrac{395}{27}\\ \Leftrightarrow2m+1=\dfrac{79}{9}\Leftrightarrow m=\dfrac{35}{9}\left(ktm\right)\\ \Leftrightarrow m\in\varnothing\)
Vậy ko có m thỏa đề
a: Để phương trình có hai nghiệm trái dấu thì 2m+1<0
hay m<-1/2
