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1.(2025·东城区期末)用配方法解方程$x^2 -8x -4=0$，变形后结果正确的是（
A.$(x-4)^2=20$
B.$(x-4)^2=16$
C.$(x-4)^2=12$
D.$(x-4)^2=4$
A
）A.$(x-4)^2=20$
B.$(x-4)^2=16$
C.$(x-4)^2=12$
D.$(x-4)^2=4$
答案
1.A
2.(2025·句容期末)若一元二次方程$x^2+6x+3=0$经过配方，变形为$(x+3)^2=m$的形式，则m的值为 （
A.0
B.3
C.6
D.9
C
）A.0
B.3
C.6
D.9
答案
2.C
3.用配方法使下列等式成立：
(1)$x^2 - 2x - 3 = (x - \_\_\_\_\_\_)^2 + (\_\_\_\_\_\_)$；
(2)$3x^2 + 2x - 2 = 3(x + \_\_\_\_\_\_)^2 + (\_\_\_\_\_\_)$。
(1)$x^2 - 2x - 3 = (x - \_\_\_\_\_\_)^2 + (\_\_\_\_\_\_)$；
(2)$3x^2 + 2x - 2 = 3(x + \_\_\_\_\_\_)^2 + (\_\_\_\_\_\_)$。
答案
3.(1)1 -4 (2)$\frac{1}{3}$ $-\frac{7}{3}$
4.填空：
(1)$x^2 + 4x + (\_\_\_\_\_\_) = (x + \_\_\_\_\_\_)^2$；
(2)$x^2 + (\_\_\_\_\_\_)x + \frac{25}{4} = (x - \frac{5}{2})^2$；
(3)$x^2 - \frac{4}{3}x + (\_\_\_\_\_\_) = (x - \_\_\_\_\_\_)^2$；
(4)$x^2 + px + (\_\_\_\_\_\_) = (x + \_\_\_\_\_\_)^2$。
(1)$x^2 + 4x + (\_\_\_\_\_\_) = (x + \_\_\_\_\_\_)^2$；
(2)$x^2 + (\_\_\_\_\_\_)x + \frac{25}{4} = (x - \frac{5}{2})^2$；
(3)$x^2 - \frac{4}{3}x + (\_\_\_\_\_\_) = (x - \_\_\_\_\_\_)^2$；
(4)$x^2 + px + (\_\_\_\_\_\_) = (x + \_\_\_\_\_\_)^2$。
答案
4.(1)4 2 (2)-5 (3)$\frac{4}{9}$ $\frac{2}{3}$ (4)$\frac{p^2}{4}$ $\frac{p}{2}$
5. 对方程$x^2+\frac{2}{5}x-\frac{3}{5}=0$进行配方，得$x^2+\frac{2}{5}x+m=\frac{3}{5}+m$，其中$m=$
$\frac{1}{25}$
.答案
5.$\frac{1}{25}$
6.当$k=$
$\pm2\sqrt{3}$
时，代数式$x^2 -kx +3$为完全平方式。答案
6.$\pm2\sqrt{3}$
7.用配方法解下列方程：
(1)$x^2 -6x -16=0$；
(2)$2x^2 -4x -1=0$；
(3)$3x^2 -x -1=0$.
(1)$x^2 -6x -16=0$；
(2)$2x^2 -4x -1=0$；
(3)$3x^2 -x -1=0$.
答案
7.解:(1)$x^2 -6x =16$,$x^2 -6x +9 =25$,$(x-3)^2 =25$,
$x-3=\pm5$,
$\therefore x_1=8$,$x_2=-2$.
(2)$2x^2 -4x =1$,$x^2 -2x =\frac{1}{2}$,$x^2 -2x +1 =\frac{3}{2}$,
即$(x-1)^2 =\frac{3}{2}$,$x-1 =\pm\frac{\sqrt{6}}{2}$,
$\therefore x_1=\frac{2+\sqrt{6}}{2}$,$x_2=\frac{2-\sqrt{6}}{2}$.
(3)$x^2 -\frac{1}{3}x -\frac{1}{3}=0$,$x^2 -\frac{1}{3}x =\frac{1}{3}$,$x^2 -\frac{1}{3}x +(\frac{1}{6})^2 =\frac{1}{3} +(\frac{1}{6})^2$,$(x-\frac{1}{6})^2 =\frac{13}{36}$,$x-\frac{1}{6} =\pm\frac{\sqrt{13}}{6}$,
$\therefore x_1=\frac{\sqrt{13}+1}{6}$,$x_2=\frac{1-\sqrt{13}}{6}$.
$x-3=\pm5$,
$\therefore x_1=8$,$x_2=-2$.
(2)$2x^2 -4x =1$,$x^2 -2x =\frac{1}{2}$,$x^2 -2x +1 =\frac{3}{2}$,
即$(x-1)^2 =\frac{3}{2}$,$x-1 =\pm\frac{\sqrt{6}}{2}$,
$\therefore x_1=\frac{2+\sqrt{6}}{2}$,$x_2=\frac{2-\sqrt{6}}{2}$.
(3)$x^2 -\frac{1}{3}x -\frac{1}{3}=0$,$x^2 -\frac{1}{3}x =\frac{1}{3}$,$x^2 -\frac{1}{3}x +(\frac{1}{6})^2 =\frac{1}{3} +(\frac{1}{6})^2$,$(x-\frac{1}{6})^2 =\frac{13}{36}$,$x-\frac{1}{6} =\pm\frac{\sqrt{13}}{6}$,
$\therefore x_1=\frac{\sqrt{13}+1}{6}$,$x_2=\frac{1-\sqrt{13}}{6}$.
8.(2025·牟平区期末)把一元二次方程$x^2+8x-8=0$化成$(x+b)^2=c$的形式，则$\sqrt{bc}$的值为（
A.$4\sqrt{6}$
B.$4\sqrt{3}$
C.$6\sqrt{3}$
D.$6\sqrt{6}$
A
）A.$4\sqrt{6}$
B.$4\sqrt{3}$
C.$6\sqrt{3}$
D.$6\sqrt{6}$
答案
8.A
9.若$△ ABC$的三边长分别为$a,b,c$,其中$a,b$满足$\sqrt{a-3}+b^2-4b+4=0$,则$c$的取值范围为
$1＜c＜5$
.答案
9.$1＜c＜5$
10.若方程$x^2 - 4100625 = 0$的两个根分别为$x_1=2025$，$x_2=-2025$，则方程$x^2 - 2x - 4100624 = 0$的两个根分别为
$x_1=2026,x_2=-2024$
.答案
10.$x_1=2026$,$x_2=-2024$
11.解下列关于$x$的方程：
(1)$x^2 + \frac{1}{6}x - \frac{1}{3}=0$；
(2)$2x^2 -7x +6=0$；
(3)$(x-1)(x-3)=7$；
(4)$x^2 -2\sqrt{5}x=4$。
(1)$x^2 + \frac{1}{6}x - \frac{1}{3}=0$；
(2)$2x^2 -7x +6=0$；
(3)$(x-1)(x-3)=7$；
(4)$x^2 -2\sqrt{5}x=4$。
答案
11.解:(1)$x^2 +\frac{1}{6}x =\frac{1}{3}$,$x^2 +\frac{1}{6}x +\frac{1}{144} =\frac{1}{3} +\frac{1}{144}$,
即$(x+\frac{1}{12})^2 =\frac{49}{144}$,$x+\frac{1}{12} =\pm\frac{7}{12}$,
$\therefore x_1=\frac{1}{2}$,$x_2=-\frac{2}{3}$.
(2)$x^2 -\frac{7}{2}x +3=0$,$(x-\frac{7}{4})^2 =\frac{1}{16}$,$x-\frac{7}{4} =\pm\frac{1}{4}$,
$\therefore x_1=2$,$x_2=\frac{3}{2}$.
(3)$x^2 -4x =4$,$x^2 -4x +4 =4+4$,$(x-2)^2 =8$,$x-2=\pm2\sqrt{2}$,$\therefore x_1=2+2\sqrt{2}$,$x_2=2-2\sqrt{2}$.
(4)$x^2 -2\sqrt{5}x +5 =4+5$,即$(x-\sqrt{5})^2 =9$,
$x-\sqrt{5} =\pm3$,$\therefore x_1=3+\sqrt{5}$,$x_2=-3+\sqrt{5}$.
即$(x+\frac{1}{12})^2 =\frac{49}{144}$,$x+\frac{1}{12} =\pm\frac{7}{12}$,
$\therefore x_1=\frac{1}{2}$,$x_2=-\frac{2}{3}$.
(2)$x^2 -\frac{7}{2}x +3=0$,$(x-\frac{7}{4})^2 =\frac{1}{16}$,$x-\frac{7}{4} =\pm\frac{1}{4}$,
$\therefore x_1=2$,$x_2=\frac{3}{2}$.
(3)$x^2 -4x =4$,$x^2 -4x +4 =4+4$,$(x-2)^2 =8$,$x-2=\pm2\sqrt{2}$,$\therefore x_1=2+2\sqrt{2}$,$x_2=2-2\sqrt{2}$.
(4)$x^2 -2\sqrt{5}x +5 =4+5$,即$(x-\sqrt{5})^2 =9$,
$x-\sqrt{5} =\pm3$,$\therefore x_1=3+\sqrt{5}$,$x_2=-3+\sqrt{5}$.
