\(3x^2-\left(3m-2\right)x-\left(3m+1\right)=0\)
có \(\Delta=\left[-\left(3m-2\right)\right]^2-4.3.\left[-\left(3m+1\right)\right]\)
\(\Delta=9m^2-12m+4+36m+12\)
\(\Delta=9m^2+24m+16\)
\(\Delta=\left(3m+4\right)^2\ge0\forall m\)
vì theo đề bài để pt có 2 nghiệm nên thỏa mãn đk \(\forall m\)
ta có vi - ét \({\begin{cases}x_1+x_2=\frac{3m-2}{3}\left(1\right)\\x_1.x_2=-\frac{\left(3m+1\right)}{3}\left(2\right)\end{cases}}\)
theo bài ra \(3x_1-5x_2=6\) \(\left(3\right)\)
từ \(\left(1\right),\left(3\right)\) ta có hệ phương trình \({\begin{cases}x_1+x_2=\frac{3m-2}{3}\\3x_1-5x_2=6\end{cases}}\)
\(\Leftrightarrow{\begin{cases}x_1+x_2=m+\frac{2}{3}\\x_1-\frac{5}{3}x_2=2\end{cases}}\)
\(\Leftrightarrow{\begin{cases}\frac{8}{3}x_2=m+\frac{2}{3}-2\\x_1+x_2=m+\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow{\begin{cases}x_2=\frac{3}{8}m-\frac{1}{2}\\x_1+\frac{3}{8}m-\frac{1}{2}=m+\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow{\begin{cases}x_2=\frac{3}{8}m-\frac{1}{2}\\x_1=m-\frac{3}{8}m+\frac{2}{3}+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow{\begin{cases}x_2=\frac{3}{8}m-\frac{1}{2}\\x_1=\frac{5}{8}m+\frac{7}{6}\end{cases}}\) \(\left(4\right)\)
Thay `(4)` vào `(2)` ta được
\(\left(\frac{3}{8}m-\frac{1}{2}\right)\left(\frac{5}{8}m+\frac{7}{6}\right)=\frac{-3m-1}{3}\)
\(\Leftrightarrow\frac{15}{64}m+\frac{7}{16}-\frac{5}{16}m-\frac{7}{12}=-m-\frac{1}{3}\)
\(\Leftrightarrow\frac{-5}{64}m-\frac{7}{48}+m+\frac{1}{3}=0\)
\(\Leftrightarrow\frac{59}{64}m+\frac{3}{16}=0\)
\(\Leftrightarrow\frac{59}{64}m=\frac{-3}{16}\)
\(\Leftrightarrow m=\frac{-12}{59}\) ( TM \(\forall m\))
Vậy \(m=\frac{-12}{59}\) là giá trị cần tìm
