1. 用公式法解下列方程:
(1)$x^2+6x+5= 0$;
(2)$2x^2+1= 4x$;
(3)$3(x^2-1)= 4x-1$;
(4)$3x(x-3)= 4(x-1)(x+1)$.

(1)解：$a=1$，$b=6$，$c=5$
$\Delta =b^2 - 4ac = 6^2 - 4×1×5 = 36 - 20 = 16 ＞ 0$
$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-6 \pm \sqrt{16}}{2×1} = \frac{-6 \pm 4}{2}$
$x_1 = \frac{-6 + 4}{2} = -1$，$x_2 = \frac{-6 - 4}{2} = -5$
(2)解：移项得$2x^2 - 4x + 1 = 0$
$a=2$，$b=-4$，$c=1$
$\Delta = (-4)^2 - 4×2×1 = 16 - 8 = 8 ＞ 0$
$x = \frac{4 \pm \sqrt{8}}{2×2} = \frac{4 \pm 2\sqrt{2}}{4} = \frac{2 \pm \sqrt{2}}{2}$
$x_1 = \frac{2 + \sqrt{2}}{2}$，$x_2 = \frac{2 - \sqrt{2}}{2}$
(3)解：去括号得$3x^2 - 3 = 4x - 1$
移项得$3x^2 - 4x - 2 = 0$
$a=3$，$b=-4$，$c=-2$
$\Delta = (-4)^2 - 4×3×(-2) = 16 + 24 = 40 ＞ 0$
$x = \frac{4 \pm \sqrt{40}}{2×3} = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}$
$x_1 = \frac{2 + \sqrt{10}}{3}$，$x_2 = \frac{2 - \sqrt{10}}{3}$
(4)解：去括号得$3x^2 - 9x = 4(x^2 - 1)$
$3x^2 - 9x = 4x^2 - 4$
移项得$-x^2 - 9x + 4 = 0$，即$x^2 + 9x - 4 = 0$
$a=1$，$b=9$，$c=-4$
$\Delta = 9^2 - 4×1×(-4) = 81 + 16 = 97 ＞ 0$
$x = \frac{-9 \pm \sqrt{97}}{2×1} = \frac{-9 \pm \sqrt{97}}{2}$
$x_1 = \frac{-9 + \sqrt{97}}{2}$，$x_2 = \frac{-9 - \sqrt{97}}{2}$

2. 已知m是方程$x(x+1)-2= 2(2x+1)$的一个根,求代数式$m^2-3m$的值.

解：方程$x(x + 1)-2=2(2x + 1)$，
整理得$x^2 + x - 2 = 4x + 2$，
$x^2 - 3x - 4 = 0$。
因为$m$是方程的根，所以$m^2 - 3m - 4 = 0$，
即$m^2 - 3m = 4$。
答案：4