《2025年名校课堂九年级数学上册华师大版8河南专版》

1. 解下列方程：
(1) $(2x + 1)^2 = 9$；
(2) $3(x + 1)^2 - 108 = 0$；
(3) $x^2 - 4x + 4 = 5$。

答案： 1.解：
(1)$2x + 1 = \pm 3.\therefore2x + 1 = 3$或$2x + 1 = - 3.\therefore x_1 = 1,x_2 = - 2.(2)3(x + 1)^2 = 108,(x + 1)^2 = 36.\therefore x + 1 = \pm 6.\therefore x_1 = 5,x_2 = - 7.(3)(x - 2)^2 = 5.\therefore x - 2 = \pm\sqrt{5}.\therefore x_1 = 2 + \sqrt{5},x_2 = 2 - \sqrt{5}$.

2. 解下列方程：
(1) $x^2 - 2x - 99 = 0$；
(2) $3x^2 - 6x + 2 = 0$。

答案： 2.解：
(1)$x^2 - 2x = 99,x^2 - 2x + 1 = 99 + 1$，即$(x - 1)^2 = 100.\therefore x - 1 = \pm 10.\therefore x_1 = 11,x_2 = - 9.(2)x^2 - 2x = -\frac{2}{3},x^2 - 2x + 1 = -\frac{2}{3} + 1$，即$(x - 1)^2 = \frac{1}{3},\therefore x - 1 = \pm\frac{\sqrt{3}}{3}.\therefore x_1 = 1 + \frac{\sqrt{3}}{3},x_2 = 1 - \frac{\sqrt{3}}{3}$.

3. 解下列方程：
(1) $x(x - 2) + x - 2 = 0$；
(2) $9x^2 - (x - 1)^2 = 0$。

答案： 3.解：
(1)$(x + 1)(x - 2) = 0,\therefore x + 1 = 0$或$x - 2 = 0.\therefore x_1 = - 1,x_2 = 2.(2)(3x + x - 1)(3x - x + 1) = 0,(4x - 1)(2x + 1) = 0,\therefore4x - 1 = 0$或$2x + 1 = 0.\therefore x_1 = \frac{1}{4},x_2 = -\frac{1}{2}$.

4. 解下列方程：
(1) $2x^2 + 2x - 1 = 0$；
(2) $3x^2 - 2x - 8 = 0$。

答案： 4.解：
(1)$\because a = 2,b = 2,c = - 1,\therefore\Delta = b^2 - 4ac = 2^2 - 4×2×(-1) = 12.\therefore x = \frac{-2\pm\sqrt{12}}{2×2} = \frac{-2\pm2\sqrt{3}}{4} = \frac{-1\pm\sqrt{3}}{2}.\therefore x_1 = \frac{-1 + \sqrt{3}}{2},x_2 = \frac{-1 - \sqrt{3}}{2}$.
(2)$\because a = 3,b = - 2,c = - 8,\therefore\Delta = b^2 - 4ac = (-2)^2 - 4×3×(-8) = 100.\therefore x = \frac{2\pm\sqrt{100}}{2×3} = \frac{2\pm10}{6}\therefore x_1 = 2,x_2 = -\frac{4}{3}$.