《2025年名校课堂九年级数学上册北师大版》

1. 下列用配方法解方程$2x^{2}+4x-1= 0$的四个步骤中,开始出现错误的步骤是(B)

A. ①
B. ②
C. ③
D. ④

答案： B

2. 用配方法解方程$3x^{2}+2x-1= 0$,配方后的方程是(D)
A. $3(x-1)^{2}= 0$
B. $(x+\frac {2}{3})^{2}= \frac {1}{3}$
C. $(x+\frac {1}{3})^{2}= \frac {1}{3}$
D. $(x+\frac {1}{3})^{2}= \frac {4}{9}$

答案： D

3. 用配方法解方程：$2x^{2}-3x-2= 0$．
解：两边同除以2,得$x^{2}-$$\frac{3}{2}x$$-1= 0$．
配方,得$x^{2}-\frac {3}{2}x+$$(\frac{3}{4})^{2}$$-$$(\frac{3}{4})^{2}$$-1= 0$．
移项,得$(x-$$\frac{3}{4}$$)^{2}=$$\frac{25}{16}$．
两边开平方,得$x-$$\frac{3}{4}$$=$$\pm\frac{5}{4}$．
所以$x_{1}=$2,$x_{2}=$$-\frac{1}{2}$．

答案： $\frac{3}{2}x$ $(\frac{3}{4})^{2}$ $(\frac{3}{4})^{2}$ $\frac{3}{4}$ $\frac{25}{16}$ $\frac{3}{4}$ $\pm\frac{5}{4}$ 2 $-\frac{1}{2}$

4. 若代数式$\frac {1}{2}x^{2}-6x$的值为7,则x的值为$6 + 5\sqrt{2}$或$6 - 5\sqrt{2}$．

答案： $6 + 5\sqrt{2}$或$6 - 5\sqrt{2}$

5. 解方程：
(1)$4x^{2}+8x= 1$．
解：$x^{2} + 2x = \frac{1}{4}$，$x^{2} + 2x + 1 = \frac{1}{4} + 1$，即$(x + 1)^{2} = \frac{5}{4}$，$\therefore x + 1 = \pm\frac{\sqrt{5}}{2}$。$\therefore x_{1} = -1 + \frac{\sqrt{5}}{2}$，$x_{2} = -1 - \frac{\sqrt{5}}{2}$
(2)$4x^{2}+4x-3= 0$．
解：$x^{2} + x = \frac{3}{4}$，$x^{2} + x + (\frac{1}{2})^{2} = \frac{3}{4} + (\frac{1}{2})^{2}$，即$(x + \frac{1}{2})^{2} = 1$，$\therefore x + \frac{1}{2} = \pm1$。$\therefore x_{1} = \frac{1}{2}$，$x_{2} = -\frac{3}{2}$
(3)$2x^{2}-1= -\frac {1}{2}x$．
解：$x^{2} + \frac{1}{4}x = \frac{1}{2}$，$x^{2} + \frac{1}{4}x + (\frac{1}{8})^{2} = \frac{1}{2} + (\frac{1}{8})^{2}$，即$(x + \frac{1}{8})^{2} = \frac{33}{64}$，$\therefore x + \frac{1}{8} = \pm\frac{\sqrt{33}}{8}$。$\therefore x_{1} = \frac{-1 + \sqrt{33}}{8}$，$x_{2} = \frac{-1 - \sqrt{33}}{8}$
(4)$-3x^{2}+4x+1= 0$．
解：$x^{2} - \frac{4}{3}x = \frac{1}{3}$，$x^{2} - \frac{4}{3}x + (\frac{2}{3})^{2} = \frac{1}{3} + (\frac{2}{3})^{2}$，即$(x - \frac{2}{3})^{2} = \frac{7}{9}$，$\therefore x - \frac{2}{3} = \pm\frac{\sqrt{7}}{3}$。$\therefore x_{1} = \frac{2 + \sqrt{7}}{3}$，$x_{2} = \frac{2 - \sqrt{7}}{3}$

答案： 解：
(1)$x^{2} + 2x = \frac{1}{4}$，$x^{2} + 2x + 1 = \frac{1}{4} + 1$，即$(x + 1)^{2} = \frac{5}{4}$，$\therefore x + 1 = \pm\frac{\sqrt{5}}{2}$。$\therefore x_{1} = -1 + \frac{\sqrt{5}}{2}$，$x_{2} = -1 - \frac{\sqrt{5}}{2}$。
(2)$x^{2} + x = \frac{3}{4}$，$x^{2} + x + (\frac{1}{2})^{2} = \frac{3}{4} + (\frac{1}{2})^{2}$，即$(x + \frac{1}{2})^{2} = 1$，$\therefore x + \frac{1}{2} = \pm1$。$\therefore x_{1} = \frac{1}{2}$，$x_{2} = -\frac{3}{2}$。
(3)$x^{2} + \frac{1}{4}x = \frac{1}{2}$，$x^{2} + \frac{1}{4}x + (\frac{1}{8})^{2} = \frac{1}{2} + (\frac{1}{8})^{2}$，即$(x + \frac{1}{8})^{2} = \frac{33}{64}$，$\therefore x + \frac{1}{8} = \pm\frac{\sqrt{33}}{8}$。$\therefore x_{1} = \frac{-1 + \sqrt{33}}{8}$，$x_{2} = \frac{-1 - \sqrt{33}}{8}$。
(4)$x^{2} - \frac{4}{3}x = \frac{1}{3}$，$x^{2} - \frac{4}{3}x + (\frac{2}{3})^{2} = \frac{1}{3} + (\frac{2}{3})^{2}$，即$(x - \frac{2}{3})^{2} = \frac{7}{9}$，$\therefore x - \frac{2}{3} = \pm\frac{\sqrt{7}}{3}$。$\therefore x_{1} = \frac{2 + \sqrt{7}}{3}$，$x_{2} = \frac{2 - \sqrt{7}}{3}$。

6. 如果一个一元二次方程的二次项是$2x^{2}$,配方后整理得$(x-\frac {1}{2})^{2}= 1$,那么它的一次项和常数项分别是(C)
A. $-x,-\frac {3}{4}$
B. $-2x,-\frac {1}{2}$
C. $-2x,-\frac {3}{2}$
D. $x,-\frac {3}{2}$

答案： C