《2025年名校课堂九年级数学上册华师大版8河南专版》

5. 解方程：
(1) $x^2 - 2x = 3$；
(2) $3x^2 - \sqrt{2}x - \frac{1}{4} = 0$；
(3) $2(x + 1)^2 = x^2 - 1$；
(4) $3x^2 - 10x + 8 = 0$。

答案： 5.解：
(1)【解法一：配方法】$x^2 - 2x + 1 = 3 + 1$，即$(x - 1)^2 = 4.\therefore x - 1 = \pm2.\therefore x_1 = 3,x_2 = - 1.$【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$x + 1 = 0.\therefore x_1 = 3,x_2 = - 1.(2)\because a = 3,b = -\sqrt{2},c = -\frac{1}{4},\therefore\Delta = (-\sqrt{2})^2 - 4×3×(-\frac{1}{4}) = 5.\therefore x = \frac{-(-\sqrt{2})\pm\sqrt{5}}{2×3} = \frac{\sqrt{2}\pm\sqrt{5}}{6},\therefore x_1 = \frac{\sqrt{2} + \sqrt{5}}{6},x_2 = \frac{\sqrt{2} - \sqrt{5}}{6}$.
(3)原方程整理，得$(x + 1)[2(x + 1) - (x - 1)] = 0,(x +1)[2x + 2 - x + 1] = 0,(x + 1)(x + 3) = 0,\therefore x + 1 = 0$或$x + 3 = 0.\therefore x_1 = - 1,x_2 = - 3.(4)(x - 2)(3x - 4) = 0,\therefore x - 2 = 0$或$3x - 4 = 0.\therefore x_1 = 2,x_2 = \frac{4}{3}$.

6. 新考向 阅读理解 阅读材料：
为解方程 $(x^2 - 1)^2 - 3(x^2 - 1) = 0$，我们可以将 $x^2 - 1$ 视为一个整体，然后设 $x^2 - 1 = y$，将原方程化为 $y^2 - 3y = 0$ ①，解得 $y_1 = 0$，$y_2 = 3$。
当 $y = 0$ 时，$x^2 - 1 = 0$，$x^2 = 1$，$\therefore x = \pm 1$；
当 $y = 3$ 时，$x^2 - 1 = 3$，$x^2 = 4$，$\therefore x = \pm 2$。
$\therefore$ 原方程的解为 $x_1 = 1$，$x_2 = -1$，$x_3 = 2$，$x_4 = -2$。
解答问题：
(1) 在由原方程得到方程①的过程中，利用换元法达到了降次的目的，体现了的数学的(B)
A. 方程思想
B. 转化思想
C. 数形结合思想
D. 建模思想
(2) 利用上述材料中的方法解方程：$(x^2 + x)^2 - (x^2 + x) - 2 = 0$。

答案： 6.解：
(1)B
(2)令$x^2 + x = m$，则原方程化为$m^2 - m - 2 = 0.\therefore(m - 2)(m + 1) = 0.\therefore m - 2 = 0$或$m + 1 = 0$，解得$m = 2$或$m = - 1$.当$m = 2$时，$x^2 + x = 2$，即$x^2 + x - 2 = 0.\therefore(x + 2)(x - 1) = 0$，则$x + 2 = 0$或$x - 1 = 0$，解得$x_1 = - 2,x_2 = 1$;当$m = - 1$时，$x^2 + x = - 1$，即$x^2 + x + 1 = 0.\therefore(x + \frac{1}{2})^2 = -\frac{3}{4} ＜ 0.\therefore$此方程没有实数根.综上所述，原方程的解为$x_1 = - 2,x_2 = 1$.