《2025年名校课堂九年级数学上册北师大版》

1. 一元二次方程$x^{2}-4= 0$的根是 (B)
A. $x_{1}= x_{2}= 2$
B. $x_{1}= 2,x_{2}= -2$
C. $x_{1}= x_{2}= 4$
D. $x_{1}= 4,x_{2}= -4$

答案： B

2. 解下列方程,能用直接开平方法的是 (D)
A. $x^{2}-x= 0$
B. $x^{2}+2= 0$
C. $x^{2}+x= 1$
D. $x^{2}-3= 6$

答案： D

3. 解方程:$(x-1)^{2}= 9$.
解:两边开平方,得$x-1=$$\pm 3$,
即$x - 1 = 3$或$x - 1 = - 3$.
所以$x_{1}=$4,$x_{2}=$$- 2$.

答案： $\pm 3$ $x - 1 = 3$ $x - 1 = - 3$ 4 $- 2$

4. 用直接开平方法解下列方程:
(1)$\frac {1}{2}x^{2}= 8$.
解: $x^{2} =$16，$\therefore x_{1} =$4，$x_{2} =$-4。
(2)$(2x-1)^{2}= 49$.
解: $2x - 1 =$±7，$\therefore 2x - 1 =$7或$2x - 1 =$-7。$\therefore x_{1} =$4，$x_{2} =$-3。
(3)$4(x+1)^{2}-25= 0$.
解: $4(x + 1)^{2} =$25，$(x + 1)^{2} =$$\frac{25}{4}$，$\therefore x + 1 =$±$\frac{5}{2}$。$\therefore x + 1 =$$\frac{5}{2}$或$x + 1 =$-$\frac{5}{2}$。$\therefore x_{1} =$$\frac{3}{2}$，$x_{2} =$-$\frac{7}{2}$。

答案： 解:
(1) $x^{2} = 16$，$\therefore x_{1} = 4$，$x_{2} = - 4$。
(2) $2x - 1 = \pm 7$，$\therefore 2x - 1 = 7$或$2x - 1 = - 7$。$\therefore x_{1} = 4$，$x_{2} = - 3$。
(3) $4(x + 1)^{2} = 25$，$(x + 1)^{2} = \frac{25}{4}$，$\therefore x + 1 = \pm \frac{5}{2}$。$\therefore x + 1 = \frac{5}{2}$或$x + 1 = - \frac{5}{2}$。$\therefore x_{1} = \frac{3}{2}$，$x_{2} = - \frac{7}{2}$。

5. (教材P36“做一做”变式)填空:
(1)$x^{2}+4x+$4$=(x+$2$)^{2}$.
(2)$x^{2}-12x+$36$=(x-$6$)^{2}$.
(3)$x^{2}+5x+$$\frac{25}{4}$$=(x+$$\frac{5}{2}$$)^{2}$.
(4)$x^{2}-\frac {2}{5}x+$$\frac{1}{25}$$=(x-$$\frac{1}{5}$$)^{2}$.

答案：
(1) 4 2
(2) 36 6
(3) $\frac{25}{4}$ $\frac{5}{2}$
(4) $\frac{1}{25}$ $\frac{1}{5}$

6. (2024·德州)把多项式$x^{2}-3x+4$进行配方,结果为 (B)
A. $(x-3)^{2}-5$
B. $(x-\frac {3}{2})^{2}+\frac {7}{4}$
C. $(x-\frac {3}{2})^{2}+\frac {25}{4}$
D. $(x+\frac {3}{2})^{2}+\frac {7}{4}$

答案： B

7. 用配方法解一元二次方程$x^{2}-6x+6= 0$时可配方得 (A)
A. $(x-3)^{2}= 3$
B. $(x+3)^{2}= 3$
C. $(x+3)^{2}= 6$
D. $(x-3)^{2}= 6$

答案： A

8. 用配方法解方程:
(1)$x^{2}+8x+1= 0$.
解: $x^{2} + 8x = - 1$，$x^{2} + 8x + 4^{2} = - 1 + 4^{2}$，$(x + 4)^{2} = 15$，$\therefore x + 4 = \pm \sqrt{15}$。$\therefore x_{1} = - 4 + \sqrt{15}$，$x_{2} = - 4 - \sqrt{15}$。
(2)$x^{2}-x-\frac {1}{4}= 0$.
解: $x^{2} - x = \frac{1}{4}$，$x^{2} - x + (\frac{1}{2})^{2} = \frac{1}{4} + (\frac{1}{2})^{2}$，$(x - \frac{1}{2})^{2} = \frac{1}{2}$，$\therefore x - \frac{1}{2} = \pm \frac{\sqrt{2}}{2}$。$\therefore x_{1} = \frac{1 + \sqrt{2}}{2}$，$x_{2} = \frac{1 - \sqrt{2}}{2}$。
(3)$x^{2}+6= 10x-2$.
解: $x^{2} - 10x = - 8$，$x^{2} - 10x + 5^{2} = - 8 + 5^{2}$，$(x - 5)^{2} = 17$，$\therefore x - 5 = \pm \sqrt{17}$。$\therefore x_{1} = 5 + \sqrt{17}$，$x_{2} = 5 - \sqrt{17}$。

答案： 解:
(1) $x^{2} + 8x = - 1$，$x^{2} + 8x + 4^{2} = - 1 + 4^{2}$，$(x + 4)^{2} = 15$，$\therefore x + 4 = \pm \sqrt{15}$。$\therefore x_{1} = - 4 + \sqrt{15}$，$x_{2} = - 4 - \sqrt{15}$。
(2) $x^{2} - x = \frac{1}{4}$，$x^{2} - x + (\frac{1}{2})^{2} = \frac{1}{4} + (\frac{1}{2})^{2}$，$(x - \frac{1}{2})^{2} = \frac{1}{2}$，$\therefore x - \frac{1}{2} = \pm \frac{\sqrt{2}}{2}$。$\therefore x_{1} = \frac{1 + \sqrt{2}}{2}$，$x_{2} = \frac{1 - \sqrt{2}}{2}$。
(3) $x^{2} - 10x = - 8$，$x^{2} - 10x + 5^{2} = - 8 + 5^{2}$，$(x - 5)^{2} = 17$，$\therefore x - 5 = \pm \sqrt{17}$。$\therefore x_{1} = 5 + \sqrt{17}$，$x_{2} = 5 - \sqrt{17}$。