Để phương trình có 2 ngiệm thì: \(\Delta>0\)
\(\Rightarrow5^2-4\cdot1\cdot\left(3m-1\right)>0\)
\(\Rightarrow29-12m>0\)
\(\Rightarrow m< \dfrac{29}{12}\)
Ta có: \(x_1^3-x_2^3+3x_1x_2=75\)
\(\Rightarrow\left(x_1-x_2\right)\left(x^2_1+x_2^2+x_1x_2\right)+3x_1x_2=75\)
\(\Rightarrow\left(x_1-x_2\right)\left(x^2_1+x^2_2+2x_1x_2-x_1x_2\right)+3x_1x_2=75\)
\(\Rightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]+3x_1x_2=75\)
\(\Rightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]+3\left(x_1x_2-25\right)\) (1)
Theo định lý Vi-ét ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=3m-1\end{matrix}\right.\)
Thay vào (1) ta có:
\(\left(x_1-x_2\right)\left(26-3m\right)+3\left(3m-26\right)=0\)
\(\Rightarrow\left(x_1-x_2-3\right)\left(26-3m\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1-x_2=3\\m=\dfrac{26}{3}\left(ktm\right)\end{matrix}\right.\)
Với \(x_1-x_2=3\)
Thì: \(\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=-4\end{matrix}\right.\)
\(\Leftrightarrow-1\cdot-4=3m-1\)
\(\Leftrightarrow m=\dfrac{5}{3}\left(tm\right)\)
Vậy: ...
