1 (1) $(x-1)^2 - 49 = 0$；
(2) $(2x-1)^2 = 25$；
(3) $3(x+2)^2 = \frac{1}{3}$；
(4) $(3x-1)^2 = (x+1)^2$。

解：​$(x-1)^2=49$​​$x-1=±7$​​$x_{1}= 8 ，$​​$x_{2}=-6$​ ; 解：​$2x-1=±5$​​$x_{1}=3 ，$​​$x_{2}=-2$​ ; 解：​$(x+2)^2=\frac {1}{9}$​​$x+2=±\frac {1}{3}$​​$x_{1}=-\frac {5}{3} ，$​​$x_{2}=-\frac {7}{3}$​ ; 解：​$3x-1=±(x+1)$​​$3x-1=x+1$​或​$3x-1=-x-1$​​$x_{1}=1 ，$​​$x_{2}=0$​

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

解：​$x^2-4x+4=3$​​$ (x-2)^2=3 $​​$x-2=±\sqrt 3 $​​$x_{1}=\sqrt 3+2 ，$​​$x_{2}=-\sqrt 3+2$​ ; 解：​$x²+3x+(\frac {3}{2})²=4+(\frac {3}{2})^2$​​$(x+\frac {3}{2})^2=\frac {25}{4}$​​$x+\frac {3}{2}=±\frac {5}{2}$​​$x_{1}=1，$​​$x_{2}=-4$​ ; 解：​$x²-2x-4=0$​​$x²-2x+1=4+1$​​$(x-1)²=5$​​$x-1=± \sqrt {5}$​​$x_{1}=1+ \sqrt {5}，$​​$x_{2}=1 -\sqrt {5}$​ ; 解：​$x^2+\frac {3}{2}x=1$​​$x²+\frac {3}{2}x+(\frac {3}{4})²=1+(\frac {3}{4})²$​​$(x+\frac {3}{4})²=\frac {25}{16}$​​$x+\frac {3}{4}=±\frac {5}{4}$​​$x_{1}=\frac {1}{2}，$​​$x_{2}=-2$​

3 (1) $x^2 - 3x + 1 = 0$；
(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 = -3，$​​$c = 1$​​$b^2 - 4ac = (-3)^2 - 4×1×1 = 5＞0$​∴​$x = \frac {-b\pm \sqrt {b^2 - 4ac}}{2a} = \frac {3\pm \sqrt {5}}{2×1}$​∴​$x_{1} = \frac {3 + \sqrt {5}}{2}，$​​$x_{2} = \frac {3 - \sqrt {5}}{2}$​ ; 解：​$a = 1，$​​$b = -2\sqrt {2}，$​​$c = 2$​​$b^2 - 4ac = (-2\sqrt {2})^2 - 4×1×2 = 0$​∴​$x = \frac {-b\pm \sqrt {b^2 - 4ac}}{2a} = \frac {-(-2\sqrt {2})\pm 0}{2} = \sqrt {2}$​∴​$x_{1} = x_{2} = \sqrt {2}$​ ; 解：​$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 = -2\ \mathrm {m^2}$​​$b^2 - 4ac =\mathrm {m^2} - 4×1×(-2\ \mathrm {m^2}) = 9\ \mathrm {m^2}$​∴​$x = \frac {-b\pm \sqrt {b^2 - 4ac}}{2a} = \frac {-m\pm 3m}{2}$​∴​$x_{1} = -2m，$​​$x_{2} = m$​

类型四 用因式分解法解方程
可化为一边为0，另一边为两个一次因式的积的形式的方程优先选用因式分解法

4 (1) $5x^2 - 4x = 0$；
(2) $x(x - 6) = -4(x - 6)$；
(3) $x(2x - 5) = 4x - 10$；
(4) $4(2x + 1)^2 - 9(2x - 1)^2 = 0$。

解：​$x(5x - 4) = 0$​​$x = 0$​或​$5x - 4 = 0$​​$x_{1} = 0，$​​$x_{2} = \frac {4}{5}$​ ; 解：​$x(x - 6) + 4(x - 6) = 0$​​$(x - 6)(x + 4) = 0$​​$x - 6 = 0$​或​$x + 4 = 0$​​$x_{1} = 6，$​​$x_{2} = -4$​ ; 解：​$x(2x - 5) - 2(2x - 5) = 0$​​$(2x - 5)(x - 2) = 0$​∴​$2x - 5 = 0$​或​$x - 2 = 0$​∴​$x_{1} = \frac {5}{2}，$​​$x_{2} = 2$​ ; 解：​$[2(2x + 1) + 3(2x - 1)][2(2x + 1) - 3(2x - 1)] = 0$​​$(10x - 1)(-2x + 5) = 0$​∴​$10x - 1 = 0$​或​$-2x + 5 = 0$​∴​$x_{1} = \frac {1}{10}，$​​$x_{2} = \frac {5}{2}$​