《2025年名校课堂九年级数学全一册人教版广西专版》

4. 解下列方程：
(1)$2x^2 + 2x - 1 = 0$.
(2)$3x^2 - 2x - 8 = 0$.
方法指导：若方程没有快捷的解法，可选择用公式法求解.

答案： 4.解：
(1)$\because a = 2,b = 2,c = - 1$，$\therefore \Delta = b^2 - 4ac = 2^2 - 4 × 2 × (-1) = 12 ＞ 0$ $\therefore x = \frac{-2 \pm \sqrt{12}}{2 × 2} = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}$ $\therefore x_1 = \frac{-1 + \sqrt{3}}{2},x_2 = \frac{-1 - \sqrt{3}}{2}$。
(2)$\because a = 3,b = - 2,c = - 8$，$\therefore \Delta = b^2 - 4ac = (-2)^2 - 4 × 3 × (-8) = 100 ＞ 0$ $\therefore x = \frac{2 \pm \sqrt{100}}{2 × 3} = \frac{2 \pm 10}{6}$ $\therefore x_1 = 2,x_2 = - \frac{4}{3}$。

5. 解下列方程：
(1)$x^2 - 2x = 3$.
(2)$9x^2 - (x - 1)^2 = 0$.
(3)$3x^2 - \sqrt{2}x - \frac{1}{4} = 0$.
(4)$3x^2 - 6x = 4(x - 2)$.

答案： 5.解：
(1)【解法一：配方法】$x^2 - 2x + 1 = 3 + 1$，即$(x - 1)^2 = 4$ $\therefore x - 1 = \pm 2$ $\therefore x_1 = 3,x_2 = - 1$。【解法二：十字相乘法】$x^2 - 2x = 3$，$x^2 - 2x - 3 = 0$，$(x - 3)(x + 1) = 0$，$\therefore x - 3 = 0$或$x + 1 = 0$ $\therefore x_1 = 3,x_2 = - 1$。
(2)$(3x + x - 1)(3x - x + 1) = 0$，$(4x - 1)(2x + 1) = 0$ $\therefore 4x - 1 = 0$或$2x + 1 = 0$ $\therefore x_1 = \frac{1}{4},x_2 = - \frac{1}{2}$。
(3)$\because a = 3,b = - \sqrt{2},c = - \frac{1}{4}$，$\therefore \Delta = (- \sqrt{2})^2 - 4 × 3 × (- \frac{1}{4}) = 5 ＞ 0$ $\therefore x = \frac{-(-\sqrt{2}) \pm \sqrt{5}}{2 × 3} = \frac{\sqrt{2} \pm \sqrt{5}}{6}$ $\therefore x_1 = \frac{\sqrt{2} + \sqrt{5}}{6},x_2 = \frac{\sqrt{2} - \sqrt{5}}{6}$。
(4)整理，得$3x^2 - 10x + 8 = 0$，$(x - 2)(3x - 4) = 0$ $\therefore x - 2 = 0$或$3x - 4 = 0$，$\therefore x_1 = 2,x_2 = \frac{4}{3}$。

6. 新考向 阅读理解(2024·桂林灌阳县期中节选)解方程$x^4 - 5x^2 + 4 = 0$，这是一个一元四次方程，根据该方程的特点，它的解法通常如下：
设$x^2 = y$，则$x^4 = y^2$，于是原方程可变为$y^2 - 5y + 4 = 0$①，解得$y_1 = 1$，$y_2 = 4$.
当$y_1 = 1$时，$x^2 = 1$，$x = \pm 1$；
当$y_2 = 4$时，$x^2 = 4$，$x = \pm 2$.
∴原方程有四个根：$x_1 = 1$，$x_2 = -1$，$x_3 = 2$，$x_4 = -2$.
(1)①中填写的方程是$y^2 - 5y + 4 = 0$，在由原方程得到方程①的过程中，利用换元法达到降次的目的，体现了数学的转化思想.
(2)参考以上方法解方程：$(x^2 + x)^2 - 4(x^2 + x) - 12 = 0$.

答案： 6.解：
(1)$y^2 - 5y + 4 = 0$ 换元
(2)设$x^2 + x = y$，则原方程可化为$y^2 - 4y - 12 = 0$。解得$y_1 = 6,y_2 = - 2$。当$y = 6$时，$x^2 + x = 6$，即$x^2 + x - 6 = 0$，解得$x_1 = - 3,x_2 = 2$；当$y = - 2$时，$x^2 + x = - 2$，即$x^2 + x + 2 = 0$。$\because \Delta = 1^2 - 4 × 1 × 2 = - 7 ＜ 0$，$\therefore$此方程无实数根。综上所述，原方程的解为$x_1 = - 3,x_2 = 2$。