《2025年名师帮同步学案九年级数学全一册人教版》

1. 因式分解法: 利用因式分解求解一元二次方程的方法叫做因式分解法.
基本原理: 左边为整式的乘积的形式, 右边等于0, 则让左边的整式分别等于0即可求解, 即: $(ax + b)(mx + n) = 0$, 则$ax + b = 0$或$mx + n = 0$.
2. 因式分解的方法: ①提公因式法; ②公式法; ③十字相乘法.

答案： 1.0 $ax + b = 0$ $mx + n = 0$
2.公式法

1. (2025·揭阳期末)解这个方程$x(2x + 3) - 3(2x + 3) = 0$最简单的方法是 (B)
A. 公式法
B. 因式分解法
C. 配方法
D. 直接开平方法

答案： 1.B

2. (2025·清远期末)$x^{2} + 5x = 0$的解为 (C)
A. $x = 0$
B. $x = -5$
C. $x_{1} = 0, x_{2} = -5$
D. $x_{1} = 0, x_{2} = 5$

答案： 2.C

3. 用因式分解法解下列方程:
(1)$x^{2} + 2x = 0$; 解:方程左边分解因式,得$x(x + 2) = 0$,$\therefore x = 0$或$x + 2 = 0$,解得$x_1 = 0$,$x_2 = -2$
(2)$2y(y - 1) - 6(y - 1) = 0$; 解:方程左边分解因式,得$(y - 1)(2y - 6) = 0$,$\therefore y - 1 = 0$或$2y - 6 = 0$,得$y_1 = 1$,$y_2 = 3$
(3)$x^{2} - 64 = 0$; 解:方程左边分解因式,得$(x + 8)(x - 8) = 0$,$\therefore x + 8 = 0$或$x - 8 = 0$,解得$x_1 = 8$,$x_2 = -8$
(4)$4x^{2} - 9 = 0$; 解:二次项系数化为1,得$x^2 - \frac{9}{4} = 0$,方程左边分解因式,得$(x + \frac{3}{2})(x - \frac{3}{2}) = 0$,$\therefore x + \frac{3}{2} = 0$或$x - \frac{3}{2} = 0$,解得$x_1 = \frac{3}{2}$,$x_2 = -\frac{3}{2}$
(5)$(x - 3)^{2} = 3(x - 3)$; 解:移项,得$(x - 3)^2 - 3(x - 3) = 0$,方程左边分解因式,得$(x - 3)(x - 3 - 3) = 0$,$\therefore x - 3 = 0$或$x - 6 = 0$,解得$x_1 = 3$,$x_2 = 6$
(6)$4(x + 3)^{2} - (x - 2)^{2} = 0$. 解:整理,得$(2x + 6)^2 - (x - 2)^2 = 0$,方程左边分解因式,得$(2x + 6 + x - 2)(2x + 6 - x + 2) = 0$,$\therefore 3x + 4 = 0$或$x + 8 = 0$,解得$x_1 = -8$,$x_2 = -\frac{4}{3}$

答案：
(1)解:方程左边分解因式,得
$x(x + 2) = 0$,
$\therefore x = 0$或$x + 2 = 0$,解得$x_1 = 0$,$x_2 = -2$;
(2)解:方程左边分解因式,得$(y - 1)(2y - 6) = 0$,
$\therefore y - 1 = 0$或$2y - 6 = 0$,得$y_1 = 1$,$y_2 = 3$;
(3)解:方程左边分解因式,得$(x + 8)(x - 8) = 0$,
$\therefore x + 8 = 0$或$x - 8 = 0$,解得$x_1 = 8$,$x_2 = -8$;
(4)解:二次项系数化为1,得$x^2 - \frac{9}{4} = 0$,
方程左边分解因式,得$(x + \frac{3}{2})(x - \frac{3}{2}) = 0$,
$\therefore x + \frac{3}{2} = 0$或$x - \frac{3}{2} = 0$,解得$x_1 = \frac{3}{2}$,
$x_2 = -\frac{3}{2}$;
(5)解:移项,得$(x - 3)^2 - 3(x - 3) = 0$,
方程左边分解因式,得$(x - 3)(x - 3 - 3) = 0$,
$\therefore x - 3 = 0$或$x - 6 = 0$,解得$x_1 = 3$,$x_2 = 6$;
(6)解:整理,得$(2x + 6)^2 - (x - 2)^2 = 0$,
方程左边分解因式,得$(2x + 6 + x - 2) - (2x + 6 - x + 2) = 0$,
$\therefore 3x + 4 = 0$或$x + 8 = 0$,解得$x_1 = -8$,$x_2 = -\frac{4}{3}$.