\(\Delta=\left(-6\right)^2-4.2.\left(3m+1\right)\)
\(=36-24m-8\)
\(=28-24m\)
Để pt có 2 nghiệm thì \(\Delta>0\)
\(\Leftrightarrow m< \dfrac{7}{6}\)
Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{6}{2}=3\\x_1.x_2=\dfrac{c}{a}=\dfrac{3m+1}{2}\end{matrix}\right.\)
Ta có: \(x_1^3+x_2^3=9\)
\(\Leftrightarrow\left(x_1+x_2\right)\left(x_1^2-x_1.x_2+x_2^2\right)=9\)
\(\Leftrightarrow\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1.x_2\right]=9\)
\(\Leftrightarrow3.\left(3^2-3.\dfrac{3m+1}{2}\right)=9\)
\(\Leftrightarrow3^2-3.\dfrac{3m+1}{2}=3\)
\(\Leftrightarrow3-\dfrac{3m+1}{2}=1\)
\(\Leftrightarrow2=\dfrac{3m+1}{2}\)
\(\Leftrightarrow3m+1=4\)
\(\Leftrightarrow3m=3\)
\(\Leftrightarrow m=1\left(tm\right)\)
Vậy `m=1` thì thỏa mãn đề bài
Phương trình đã cho có nghiệm khi:
\(\Delta'=9-2\left(3m+1\right)\ge0\Rightarrow m\le\dfrac{7}{6}\)
Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=\dfrac{3m+1}{2}\end{matrix}\right.\)
\(x_1^3+x_2^3=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=9\)
\(\Leftrightarrow3^3-3.3.\left(\dfrac{3m+1}{2}\right)=9\)
\(\Leftrightarrow m=1\) (thỏa mãn)
