1.(2025·玄武区期中)用配方法解一元二次方程$2x^{2}-4x-1=0$时,配方正确的是 (
A.$(x-1)^{2}=\dfrac{3}{2}$
B.$(x-1)^{2}=\dfrac{1}{2}$
C.$(x-2)^{2}=3$
D.$(x-2)^{2}=5$
A
)A.$(x-1)^{2}=\dfrac{3}{2}$
B.$(x-1)^{2}=\dfrac{1}{2}$
C.$(x-2)^{2}=3$
D.$(x-2)^{2}=5$
答案
A
2. 填空:(1)$2x^{2}-12x+\_\_\_\_\_\_=2(x-\_\_\_\_\_\_)^{2}$;(2)$-m^{2}+2\sqrt{3}m-\_\_\_\_\_\_=-(m-\_\_\_\_\_\_)^{2}$;
(3)$3x^{2}-12x+\_\_\_\_\_\_=3(x-2)^{2}$;
(4)$16x^{2}+12x+\_\_\_\_\_\_=16(x+\_\_\_\_\_\_)^{2}$.
(3)$3x^{2}-12x+\_\_\_\_\_\_=3(x-2)^{2}$;
(4)$16x^{2}+12x+\_\_\_\_\_\_=16(x+\_\_\_\_\_\_)^{2}$.
答案
(1)18 3 (2)3 $\sqrt{3}$ (3)12 (4)$\dfrac{9}{4}$ $\dfrac{3}{8}$
3. 用配方法解下列方程:
(1)$2x^{2}+8x-1=0$;
(2)$(x-2)^{2}=3x(x-2)$;
(3)$3y^{2}=6y+5$;
(4)$2x^{2}-2=3x$;
(5)$3x^{2}-2x-2=0$;
(6)$6x^{2}-x=12.$
(1)$2x^{2}+8x-1=0$;
(2)$(x-2)^{2}=3x(x-2)$;
(3)$3y^{2}=6y+5$;
(4)$2x^{2}-2=3x$;
(5)$3x^{2}-2x-2=0$;
(6)$6x^{2}-x=12.$
答案
(1)$x_1=-2+\dfrac{3\sqrt{2}}{2},x_2=-2-\dfrac{3\sqrt{2}}{2}$
(2)$x_1=2,x_2=-1$
(3)$y_1=\dfrac{3+2\sqrt{6}}{3},y_2=\dfrac{3-2\sqrt{6}}{3}$
(4)$x_1=2,x_2=-\dfrac{1}{2}$
(5)$x_1=\dfrac{1+\sqrt{7}}{3},x_2=\dfrac{1-\sqrt{7}}{3}$
(6)$x_1=\dfrac{3}{2},x_2=-\dfrac{4}{3}$
(2)$x_1=2,x_2=-1$
(3)$y_1=\dfrac{3+2\sqrt{6}}{3},y_2=\dfrac{3-2\sqrt{6}}{3}$
(4)$x_1=2,x_2=-\dfrac{1}{2}$
(5)$x_1=\dfrac{1+\sqrt{7}}{3},x_2=\dfrac{1-\sqrt{7}}{3}$
(6)$x_1=\dfrac{3}{2},x_2=-\dfrac{4}{3}$
4.(2025·海门区模拟)用配方法解一元二次方程$2x^{2}+4x-5=0$时,将它化为$(x+a)^{2}=b$的形式,则$a+b$的值为(
A.8
B.$\dfrac{9}{2}$
C.$\dfrac{7}{2}$
D.$\dfrac{5}{2}$
B
)A.8
B.$\dfrac{9}{2}$
C.$\dfrac{7}{2}$
D.$\dfrac{5}{2}$
答案
B
5. 用配方法解下列方程时,配方有错误的是(
A.$x^2 - 2x - 99 = 0$ 化为 $(x - 1)^2 = 100$
B.$x^2 + 8x + 9 = 0$ 化为 $(x + 4)^2 = 25$
C.$2t^2 - 7t - 4 = 0$ 化为 $(t - \dfrac{7}{4})^2 = \dfrac{81}{16}$
D.$3y^2 - 4y - 2 = 0$ 化为 $(y - \dfrac{2}{3})^2 = \dfrac{10}{9}$
B
)A.$x^2 - 2x - 99 = 0$ 化为 $(x - 1)^2 = 100$
B.$x^2 + 8x + 9 = 0$ 化为 $(x + 4)^2 = 25$
C.$2t^2 - 7t - 4 = 0$ 化为 $(t - \dfrac{7}{4})^2 = \dfrac{81}{16}$
D.$3y^2 - 4y - 2 = 0$ 化为 $(y - \dfrac{2}{3})^2 = \dfrac{10}{9}$
答案
B
6. 下列用配方法解方程$\dfrac{1}{2}x^{2}-x-2=0$的四个步骤中,开始出现错误的是(
$\frac{1}{2}x^2 - x - 2 = 0 \xrightarrow{①} x^2 - 2x = 4 \xrightarrow{②} x^2 - 2x + 1 = 5 \xrightarrow{③} (x - 1)^2 = 5 \xrightarrow{④} x = \sqrt{5} + 1$
A.①
B.②
C.③
D.④
D
)$\frac{1}{2}x^2 - x - 2 = 0 \xrightarrow{①} x^2 - 2x = 4 \xrightarrow{②} x^2 - 2x + 1 = 5 \xrightarrow{③} (x - 1)^2 = 5 \xrightarrow{④} x = \sqrt{5} + 1$
A.①
B.②
C.③
D.④
答案
D
7. 已知等腰三角形的两边长 $a,b$ 满足 $4a^{2}-4ab+2b^{2}-8b+16=0$,则此等腰三角形的周长为
(
A.8
B.10
C.12
D.8 或 10
(
B
)A.8
B.10
C.12
D.8 或 10
答案
B
8. 当$x=$
1或$\dfrac{1}{2}$
时,代数式$5x^{2}-2x-1$与$3x^{2}+x-2$的值相等.答案
1或$\dfrac{1}{2}$
9. 若一元二次方程$4x^{2}+12x-27=0$的两根分别为$a,b$,且$a>b$,则$3a+b$的值为
0
.答案
0
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