《2025年名校课堂九年级数学上册人教版》

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$(x - 1)^2 = 4.\therefore x - 1 = \pm 2.\therefore x_1 = 3,x_2 = - 1$．【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)$\because a = 3,b = - \sqrt{2},c = - \frac{1}{4},\therefore \Delta = ( - \sqrt{2})^2 - 4 × 3 × ( - \frac{1}{4}) = 5 ＞ 0.\therefore x = \frac{- ( - \sqrt{2}) \pm \sqrt{5}}{2 × 3} = \frac{\sqrt{2} \pm \sqrt{5}}{6},\therefore x_1 = \frac{\sqrt{2} + \sqrt{5}}{6},x_2 = \frac{\sqrt{2} - \sqrt{5}}{6}$．
(3)原程整理，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，$x^{2}-1=3,x^{2}=4,\therefore x=\pm 2.$
∴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$x^2 + x + 1 = 0,\because \Delta = 1^2 - 4 × 1 × 1 = - 3 ＜ 0$，此方程无实数根．综上所述，原方程的解为$x_1 = - 2,x_2 = 1$．