《2025年零障碍导教导学案九年级数学全一册北师大版》

1. 例(※选学)用十字相乘法解方程:
(1)$x^{2}-8x+12=0$; 解:$(x - 2)(x - 6) = 0$.$x - 2 = 0$,或$x - 6 = 0$.$x_1 = 2$,$x_2 = 6$.(2)$x^{2}+7x+12=0$.解:$(x + 3)(x + 4) = 0$.$x + 3 = 0$,或$x + 4 = 0$.$x_1 = -3$,$x_2 = -4$.

答案：
(1)解:$(x - 2)(x - 6) = 0$.
$x - 2 = 0$,或$x - 6 = 0$.
$x_1 = 2$,$x_2 = 6$.
(2)解:$(x + 3)(x + 4) = 0$.
$x + 3 = 0$,或$x + 4 = 0$.
$x_1 = -3$,$x_2 = -4$.

2. 用十字相乘法解方程:
(1)$x^{2}-4x-12=0$; 解:$(x - 6)(x + 2) = 0$.$x - 6 = 0$,或$x + 2 = 0$.$x_1 = 6$,$x_2 = -2$.
(2)$x^{2}+x-12=0$.解:$(x + 4)(x - 3) = 0$.$x + 4 = 0$,或$x - 3 = 0$.$x_1 = -4$,$x_2 = 3$.

答案：
(1)解:$(x - 6)(x + 2) = 0$.
$x - 6 = 0$,或$x + 2 = 0$.
$x_1 = 6$,$x_2 = -2$.
(2)解:$(x + 4)(x - 3) = 0$.
$x + 4 = 0$,或$x - 3 = 0$.
$x_1 = -4$,$x_2 = 3$.

3. 例(2024·南海区模拟)解方程:$2x^{2}+3x-5=0$.

答案： 解:$(2x + 5)(x - 1) = 0$,
$2x + 5 = 0$,或$x - 1 = 0$.
$x_1 = -\frac{5}{2}$,$x_2 = 1$.

4. (2024·深圳模拟)解方程:$3x^{2}-7x+2=0$.

答案： 解:$(3x - 1)(x - 2) = 0$,
$3x - 1 = 0$,或$x - 2 = 0$.
$x_1 = \frac{1}{3}$,$x_2 = 2$.

5. 用直接开平方法解方程:$x^{2}-4=0$.

答案： 解:$x^2 = 4$.
$x = \pm\sqrt{4}$.
$x_1 = 2$,$x_2 = -2$.

6. (2024·福田区月考改编)用因式分解法解方程:$\frac {1}{2}x=x^{2}$.

答案： 解:原方程可变形为
$x(x - \frac{1}{2}) = 0$,
$x = 0$,或$x - \frac{1}{2} = 0$.
$x_1 = 0$,$x_2 = \frac{1}{2}$.

7. (2024·顺德区模拟)用公式法解方程:$2x^{2}+3x-1=0$.

答案： 解:$\because a = 2$,$b = 3$,$c = -1$,
$\Delta = b^2 - 4ac$
$= 3^2 - 4\times2\times(-1)$
$= 17 ＞ 0$,
$\therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$= \frac{-3 \pm \sqrt{17}}{2\times2}$
$= \frac{-3 \pm \sqrt{17}}{4}$.
$\therefore x_1 = \frac{-3 + \sqrt{17}}{4}$,
$x_2 = \frac{-3 - \sqrt{17}}{4}$.

8. (2024·梅州期中)用配方法解方程:$x^{2}-2\sqrt {3}x=3$.

答案： 解:$x^2 - 2\sqrt{3}x + 3 = 3 + 3$.
$(x - \sqrt{3})^2 = 6$.
$x - \sqrt{3} = \pm\sqrt{6}$.
$x = \sqrt{3} \pm\sqrt{6}$.
$\therefore x_1 = \sqrt{3} + \sqrt{6}$,$x_2 = \sqrt{3} - \sqrt{6}$.