《2025年名校课堂九年级数学上册北师大版》

5. 解下列方程:
(1)$x^{2}-2x= 3$.
【解法一: 配方法】$x^{2} - 2x + 1 = 3 + 1$,$(x - 1)^{2} = 4$,$\therefore x - 1 = \pm 2$.$\therefore x_{1} = 3$,$x_{2} = - 1$.【解法二: 十字相乘法】$x^{2} - 2x - 3 = 0$,$(x - 3)(x + 1) = 0$,$\therefore x_{1} = 3$,$x_{2} = - 1$.
(2)$3x^{2}-\sqrt {2}x-\frac {1}{4}= 0$.
$\because a = 3$,$b = - \sqrt{2}$,$c = - \frac{1}{4}$,$\therefore \Delta = (- \sqrt{2})^{2} - 4 × 3 × (- \frac{1}{4}) = 5 ＞ 0$.$\therefore x = \frac{- (- \sqrt{2}) \pm \sqrt{5}}{2 × 3}$.$\therefore x_{1} = \frac{\sqrt{2} + \sqrt{5}}{6}$,$x_{2} = \frac{\sqrt{2} - \sqrt{5}}{6}$.
(3)$2(x+1)^{2}= x^{2}-1$.
$2(x + 1)^{2} - (x + 1)(x - 1) = 0$,$(x + 1)[2(x + 1) - (x - 1)] = 0$,$(x + 1)(x + 3) = 0$,$\therefore x + 1 = 0$或$x + 3 = 0$.$\therefore x_{1} = - 1$,$x_{2} = - 3$.
(4)$3x^{2}-10x+8= 0$.
$(x - 2)(3x - 4) = 0$,$\therefore x - 2 = 0$或$3x - 4 = 0$.$\therefore x_{1} = 2$,$x_{2} = \frac{4}{3}$.

答案： 解:
(1)【解法一: 配方法】$x^{2} - 2x + 1 = 3 + 1$,$(x - 1)^{2} = 4$,$\therefore x - 1 = \pm 2$.$\therefore x_{1} = 3$,$x_{2} = - 1$.【解法二: 十字相乘法】$x^{2} - 2x - 3 = 0$,$(x - 3)(x + 1) = 0$,$\therefore x_{1} = 3$,$x_{2} = - 1$.
(2)$\because a = 3$,$b = - \sqrt{2}$,$c = - \frac{1}{4}$,$\therefore \Delta = (- \sqrt{2})^{2} - 4 \times 3 \times (- \frac{1}{4}) = 5 ＞ 0$.$\therefore x = \frac{- (- \sqrt{2}) \pm \sqrt{5}}{2 \times 3}$.$\therefore x_{1} = \frac{\sqrt{2} + \sqrt{5}}{6}$,$x_{2} = \frac{\sqrt{2} - \sqrt{5}}{6}$.
(3)$2(x + 1)^{2} - (x + 1)(x - 1) = 0$,$(x + 1)[2(x + 1) - (x - 1)] = 0$,$(x + 1)(x + 3) = 0$,$\therefore x + 1 = 0$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:$(x^{2}+x)^{2}-(x^{2}+x)-2= 0$.
解: 设$x^{2} + x = m$,则$m^{2} - m - 2 = 0$.$\therefore (m - 2)(m + 1) = 0$.$\therefore m - 2 = 0$或$m + 1 = 0$,解得$m = 2$或$m = - 1$. ①当$m = 2$时,$x^{2} + x = 2$,即$x^{2} + x - 2 = 0$,$\therefore (x + 2)(x - 1) = 0$.$\therefore x + 2 = 0$或$x - 1 = 0$,解得$x_{1} = - 2$,$x_{2} = 1$; ②当$m = - 1$时,$x^{2} + x = - 1$,即$x^{2} + x + 1 = 0$,$\because \Delta = 1^{2} - 4 × 1 × 1 = - 3 ＜ 0$,$\therefore$此方程没有实数根. 综上所述,原方程的解为$x_{1} = - 2$,$x_{2} = 1$.

答案： 解:
(1)B
(2)设$x^{2} + x = m$,则$m^{2} - m - 2 = 0$.$\therefore (m - 2)(m + 1) = 0$.$\therefore m - 2 = 0$或$m + 1 = 0$,解得$m = 2$或$m = - 1$. ①当$m = 2$时,$x^{2} + x = 2$,即$x^{2} + x - 2 = 0$,$\therefore (x + 2)(x - 1) = 0$.$\therefore x + 2 = 0$或$x - 1 = 0$,解得$x_{1} = - 2$,$x_{2} = 1$; ②当$m = - 1$时,$x^{2} + x = - 1$,即$x^{2} + x + 1 = 0$,$\because \Delta = 1^{2} - 4 \times 1 \times 1 = - 3 ＜ 0$,$\therefore$此方程没有实数根. 综上所述,原方程的解为$x_{1} = - 2$,$x_{2} = 1$.