\(3x^2+6x+m=0\)
\(\Delta'=9-3m\)
Để pt có nghiệm<=> \(\Delta'\ge0\Leftrightarrow m\le3\)
Theo Vi-ét:
\(\left\{{}\begin{matrix}x_1+x_2=-\frac{6}{3}=-2\\x_1.x_2=\frac{m}{3}\end{matrix}\right.\)
\(\left|x_1-x_2\right|=1\Leftrightarrow\left(x_1-x_2\right)^2=1\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=1\)
\(\Leftrightarrow4-\frac{4m}{3}=1\)
\(\Leftrightarrow m=\frac{9}{4}\left(tm\right)\)