《2025年名师帮同步学案九年级数学全一册人教版》

【例题1】用十字相乘法解下列方程：
(1)$x^{2}+6x+8 = 0$；
解:$(x + 2)(x + 4) = 0$，
$\therefore x + 2 = 0$或$x + 4 = 0$，
$\therefore x_1 = $$-2$，$x_2 = $$-4$；
(2)$x^{2}-12x+35 = 0$；
解:$(x - 5)(x - 7) = 0$，
$\therefore x - 5 = 0$或$x - 7 = 0$，
$\therefore x_1 = $$5$，$x_2 = $$7$。

答案： 解:
(1)$(x + 2)(x + 4) = 0$，
$\therefore x + 2 = 0$或$x + 4 = 0$，
$\therefore x_1 = -2$，$x_2 = -4$；
(2)$(x - 5)(x - 7) = 0$，
$\therefore x - 5 = 0$或$x - 7 = 0$，
$\therefore x_1 = 5$，$x_2 = 7$。

【变式1】用十字相乘法解下列方程：
(1)$x^{2}-x-20 = 0$；
解: $(x - 5)(x + 4) = 0$，
$\therefore x - 5 = 0$或$x + 4 = 0$，
$\therefore x_1 = $5，$x_2 = $-4；
(2)$x^{2}+7x-8 = 0$。
解: $(x - 1)(x + 8) = 0$，
$\therefore x - 1 = 0$或$x + 8 = 0$，
$\therefore x_1 = $1，$x_2 = $-8。

答案： 解:
(1)$(x - 5)(x + 4) = 0$，
$\therefore x - 5 = 0$或$x + 4 = 0$，
$\therefore x_1 = 5$，$x_2 = -4$；
(2)$(x - 1)(x + 8) = 0$，
$\therefore x - 1 = 0$或$x + 8 = 0$，
$\therefore x_1 = 1$，$x_2 = -8$。

【例题2】用十字相乘法解下列方程：
(1)$12x^{2}+7x+1 = 0$；
解:$(4x + 1)(3x + 1) = 0$，
$\therefore 4x + 1 = 0$或$3x + 1 = 0$，
$\therefore x_1 = $$-\frac{1}{4}$，$x_2 = $$-\frac{1}{3}$；
(2)$3x^{2}-11x+6 = 0$。
解:$(3x - 2)(x - 3) = 0$，
$\therefore 3x - 2 = 0$或$x - 3 = 0$，
$\therefore x_1 = $$\frac{2}{3}$，$x_2 = $$3$。

答案： 解:
(1)$(4x + 1)(3x + 1) = 0$，
$\therefore 4x + 1 = 0$或$3x + 1 = 0$，
$\therefore x_1 = -\frac{1}{4}$，$x_2 = -\frac{1}{3}$；
(2)$(3x - 2)(x - 3) = 0$，
$\therefore 3x - 2 = 0$或$x - 3 = 0$，
$\therefore x_1 = \frac{2}{3}$，$x_2 = 3$。

【变式2】用十字相乘法解下列方程：
(1)$2x^{2}+3x+1 = 0$；
解:$(x + 1)(2x + 1) = 0$，
$\therefore x + 1 = 0$或$2x + 1 = 0$，
$\therefore x_1 = $$-1$，$x_2 = $$-\frac{1}{2}$；
(2)$2x^{2}-5x-3 = 0$；
解:$(2x + 1)(x - 3) = 0$，
$\therefore 2x + 1 = 0$或$x - 3 = 0$，
$\therefore x_1 = $$-\frac{1}{2}$，$x_2 = $$3$。

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

【例题3】解方程：$(40 - x)(20 + 2x) = 1200$。

答案： 解: 整理，得$x^2 - 30x + 200 = 0$，
$(x - 10)(x - 20) = 0$，
$\therefore x - 10 = 0$或$x - 20 = 0$，
$\therefore x_1 = 10$，$x_2 = 20$。

【变式3】解方程：$\frac{1}{2}x(x - 1) = 55$。

答案： 解: 整理，得$x^2 - x - 110 = 0$，
$(x - 11)(x + 10) = 0$，
$\therefore x - 11 = 0$或$x + 10 = 0$，
解得$x_1 = 11$，$x_2 = -10$。