證明??$：(1)△=(2m-1)2-4×1×(-3m2+m)$??
??$=4m2-4m+1+12m2-4m$??
??$=16m2-8m+1$??
??$=(4m-1)2≥0$??
∴無論??$m$??為何值,方程總有實數(shù)根。
??$(2)$??根據(jù)題意，得??$x_{1}+x_{2}=2m-1，x_{1}x_{2}=-3m2+m. $??
??$∵\frac {x_2}{x_1}+\frac {x_1}{x_2}$??
??$=\frac {x_12+x_22}{x_1x_2}$??
??$=\frac {(x_1+x_2)2-2x_1x_2}{x_1x_2}$??
??$=\frac {(x_1+x_2)2}{x_1x_2}-2$??
??$=-\frac {5}{2}，$??
??$∴ \frac {(2m-1)2}{3m2+m}-2=-\frac {5}{2}. $??
整理，得??$5m2-7m+2=0，$??
解得??$m_{1}=1，m_{2}=\frac {2}{5}$??
??$∴m $??的值為??$1$??或??$\frac {2}{5}$??
