1. (1)$(x-3)^2 - 4 = 0$;
(2)$(2x-1)^2 = 25$;
(3)$3(x+2)^2 = \frac{1}{3}$;
(4)$(3x-1)^2 = (x+1)^2$.

解：$∵(x - 3)^2 - 4 = 0，$
$∴(x - 3)^2 = 4，$
$∴x - 3 = \pm2，$ $∴x_1 = 1，$$x_2 = 5。$ ; 解：$∵(2x - 1)^2 = 25，$
$∴2x - 1 = \pm5，$ $∴x_1 = 3，$$x_2 = -2。$ ; 解：$∵3(x + 2)^2 = \frac{1}{3}，$
$∴(x + 2)^2 = \frac{1}{9}，$
$∴x + 2 = \pm\frac{1}{3}，$ $∴x_1 = -\frac{5}{3}，$$x_2 = -\frac{7}{3}。$ ; 解：$∵(3x - 1)^2 = (x + 1)^2，$
$∴3x - 1 = \pm(x + 1)，$ $∴2x = 2$或$4x = 0，$
$∴x_1 = 1，$$x_2 = 0。$

2. (1)$x^2 - 4x - 1 = 0$；
(2)$x^2 + 3x - 4 = 0$；
(3)$-\dfrac{1}{2}x^2 + x + 2 = 0$；
(4)$2x^2 - 5x + 2 = 0$。

解：移项，得$x^2 - 4x = 1，$
配方，得$x^2 - 4x + 4 = 1 + 4，$ 即$(x - 2)^2 = 5，$
所以$x - 2 = \pm\sqrt{5}，$
所以$x_1 = 2 + \sqrt{5}，$$x_2 = 2 - \sqrt{5}。$ ; 解：移项，得$x^2 + 3x = 4，$ 配方，得$x^2 + 3x + (\frac{3}{2})^2 = 4 + (\frac{3}{2})^2，$ 即$(x + \frac{3}{2})^2 = \frac{25}{4}，$ 所以$x + \frac{3}{2} = \pm\frac{5}{2}，$
所以$x_1 = 1，$$x_2 = -4。$ ; 解：化二次项系数为$1，$得$x^2 - 2x - 4 = 0，$ 移项，得$x^2 - 2x = 4，$ 配方，得$x^2 - 2x + 1 = 4 + 1，$ 即$(x - 1)^2 = 5，$所以$x - 1 = \pm\sqrt{5}，$ 所以$x_1 = 1 + \sqrt{5}，$$x_2 = 1 - \sqrt{5}。$ ; 解：化二次项系数为$1，$得$x^2 - \frac{5}{2}x + 1 = 0，$
移项，得$x^2 - \frac{5}{2}x = -1，$ 配方，得$x^2 - \frac{5}{2}x + (\frac{5}{4})^2 = -1 + (\frac{5}{4})^2，$ 即$(x - \frac{5}{4})^2 = \frac{9}{16}，$
所以$x - \frac{5}{4} = \pm\frac{3}{4}，$
所以$x_1 = 2，$$x_2 = \frac{1}{2}。$

3. (1)$x^2 - 5x + 1 = 0$；
(2)$x^2 - 2\sqrt{2}x + 2 = 0$；
(3)$x(x + 1) + 4(x - 1) = 2(x - 4)$；
(4)$x^2 + mx - 2m^2 = 0$（$m$为常数）。

解：$∵a = 1，$$b = -5，$$c = 1，$ $∴b^2 - 4ac = (-5)^2 - 4×1×1 = 21＞0，$ $∴x = \frac{-(-5)\pm\sqrt{21}}{2×1}，$
$∴x_1 = \frac{5 + \sqrt{21}}{2}，$$x_2 = \frac{5 - \sqrt{21}}{2}。$ ; 解：$∵a = 1，$$b = -2\sqrt{2}，$$c = 2，$ $∴b^2 - 4ac = (-2\sqrt{2})^2 - 4×1×2 = 0，$ $∴x = \frac{-(-2\sqrt{2})\pm0}{2×1}=\sqrt{2}，$
$∴x_1 = x_2 = \sqrt{2}。$ ; 解：化方程为一般形式，得$x^2 + 3x + 4 = 0。$ $∵a = 1，$$b = 3，$$c = 4，$ $∴b^2 - 4ac = 3^2 - 4×1×4 = 9 - 16 = -7＜0，$ $∴$此方程没有实数根。 ; 解：$∵a = 1，$$b = m，$$c = -2m^2，$ $∴b^2 - 4ac = m^2 - 4×1×(-2m^2) = 9m^2，$ $∴x = \frac{-m\pm3m}{2×1}，$
$∴x_1 = -2m，$$x_2 = m。$

4. (1)$5x^2 - 4x = 0$;
(2)$x(x - 6) = -4(x - 6)$;
(3)$x^2 - 3x = x - 3$;
(4)$4(2x + 1)^2 - 9(2x - 1)^2 = 0$.

解：原方程可化为$x(5x - 4) = 0，$
所以$x = 0$或$5x - 4 = 0，$
所以$x_1 = 0，$$x_2 = \frac{4}{5}。$ ; 解：移项，得$x(x - 6) + 4(x - 6) = 0，$
即$(x - 6)(x + 4) = 0，$ 所以$x - 6 = 0$或$x + 4 = 0，$
所以$x_1 = 6，$$x_2 = -4。$ ; 解：移项，得$x(x - 3) - (x - 3) = 0，$
因式分解，得$(x - 3)(x - 1) = 0，$ 则$x - 3 = 0$或$x - 1 = 0，$
所以$x_1 = 3，$$x_2 = 1。$ ; 解：因式分解，得​$[2(2x + 1) + 3(2x - 1)][2(2x + 1) - 3(2x - 1)] = 0，$​​$ $​即​$(4x + 2 + 6x - 3)(4x + 2 - 6x + 3) = 0，$​​$(10x - 1)(-2x + 5) = 0，$​​$ $​则​$10x - 1 = 0$​或​$-2x + 5 = 0，$​所以​$x_{1} = \frac {1}{10}，$​​$x_{2} = \frac {5}{2}。$​