已知關(guān)于$x$的一元二次方程$mx^2 - (3m - 1)x + 2m - 1 = 0，$其根的判別式的值為$1。$
因?yàn)榉匠淌且辉畏匠?，所以二次?xiàng)系數(shù)$m \neq 0。$
根的判別式$\Delta = b^2 - 4ac，$其中$a = m，$$b = -(3m - 1)，$$c = 2m - 1。$
則$\Delta = [-(3m - 1)]^2 - 4m(2m - 1)$
$= (3m - 1)^2 - 4m(2m - 1)$
$= 9m^2 - 6m + 1 - 8m^2 + 4m$
$= m^2 - 2m + 1$
已知$\Delta = 1，$所以$m^2 - 2m + 1 = 1$
即$m^2 - 2m = 0$
$m(m - 2) = 0$
解得$m = 0$或$m = 2$
又因?yàn)?m \neq 0，$所以$m = 2$
將$m = 2$代入原方程得：$2x^2 - (3×2 - 1)x + 2×2 - 1 = 0$
即$2x^2 - 5x + 3 = 0$
對(duì)于方程$2x^2 - 5x + 3 = 0，$使用求根公式$x = \frac{-b \pm \sqrt{\Delta}}{2a}，$其中$a = 2，$$b = -5，$$\Delta = 1$
$x = \frac{5 \pm \sqrt{1}}{2×2} = \frac{5 \pm 1}{4}$
所以$x_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}，$$x_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1$
綜上，$m = 2，$方程的根為$x_1 = 1，$$x_2 = \frac{3}{2}。$
