《2025年全品学练考九年级数学上册人教版》

变式 解下列方程:
(1)$(x + 2)^{2} = 3x + 6$;
(2)$x^{2} - 1 = 2(x + 1)$.

答案：
(1)$ x_{1} = -2,x_{2} = 1 $
(2)$ x_{1} = -1,x_{2} = 3 $

例3(教材补充例题)选择合适的方法解下列方程:
(1)$x^{2} + x = 0$;
(2)$2x^{2} + 3x - 5 = 0$;
(3)$x^{2} - 8x - 1 = 0$;
(4)$(2x + 8)(x - 2) = x^{2} + 2x - 17$.

答案：
(1)$ x_{1} = 0,x_{2} = -1 $
(2)$ x_{1} = 1,x_{2} = -\frac{5}{2} $
(3)$ x_{1} = 4 + \sqrt{17},x_{2} = 4 - \sqrt{17} $
(4)$ x_{1} = x_{2} = -1 $

1. 用因式分解法解下列方程:
(1)$2x^{2} + x = 0$;
(2)$(x + 1)^{2} - 2(x + 1) = 0$.

答案：
(1)$ x_{1} = 0,x_{2} = -\frac{1}{2} $
(2)$ x_{1} = -1,x_{2} = 1 $

2. 选择恰当的方法解下列一元二次方程:
(1)$x^{2} = 8$;
(2)$x^{2} - 2x - 5 = 0$;
(3)$2x^{2} - 5x + 2 = 0$;
(4)$(2x - 5)^{2} - 9 = 0$.

答案：
(1)$ x_{1} = 2\sqrt{2},x_{2} = -2\sqrt{2} $
(2)$ x_{1} = 1 + \sqrt{6},x_{2} = 1 - \sqrt{6} $
(3)$ x_{1} = 2,x_{2} = \frac{1}{2} $
(4)$ x_{1} = 1,x_{2} = 4 $