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1 [2024·湖南长沙月考]把方程3x² - 12x - 18 = 0配方，化为(x + m)² = n的形式应为 ( )
A. (x - 4)² = 6
B. (x - 2)² = 4
C. (x - 2)² = 10
D. (x - 2)² = 0
A. (x - 4)² = 6
B. (x - 2)² = 4
C. (x - 2)² = 10
D. (x - 2)² = 0
答案：
C
2 用配方法解方程2x² - 12x = 5时，先把二次项系数化为1，然后方程的两边应都加上( )
A. 4
B. 9
C. 25
D. 36
A. 4
B. 9
C. 25
D. 36
答案：
B
3 链教材P35作业题第1题改编 [2024·山东菏泽期末]用配方法解方程2x² - 2x - 3 = 0时，配方后方程变形为________.
答案：
$(x - \frac{1}{2})^2 = \frac{7}{4}$
4 [2024·安徽安庆期末]把方程2x² + 8x - 1 = 0化为(x + m)² = n的形式，则√mn的值是________.
答案：
3
5 用配方法解方程：2x² - 3x - 2 = 0.
解：二次项系数化为1，移项，
得x² - 3/2x = ________.
配方，得x² - 3/2x + ________ = ________，
即(x - 3/4)² = 25/16.
方程两边同时开平方，得x - 3/4 = ________.
解得x₁ = ________，x₂ = ________.
解：二次项系数化为1，移项，
得x² - 3/2x = ________.
配方，得x² - 3/2x + ________ = ________，
即(x - 3/4)² = 25/16.
方程两边同时开平方，得x - 3/4 = ________.
解得x₁ = ________，x₂ = ________.
答案：
1 $\frac{9}{16}$ $\frac{25}{16}$ $\pm\frac{5}{4}$ 2 $-\frac{1}{2}$
6 链教材P35作业题第3题改编 用配方法解下列方程：
(1)2x² - 4x + 1 = 0;
(2)2x² - 5x + 3 = 0;
(3)1/4x² - 6x + 3 = 0.
(1)2x² - 4x + 1 = 0;
(2)2x² - 5x + 3 = 0;
(3)1/4x² - 6x + 3 = 0.
答案：
解：
(1)方程的两边同除以2，得$x^2 - 2x + \frac{1}{2} = 0$.
移项，得$x^2 - 2x = -\frac{1}{2}$.
方程的两边同加上1，得$x^2 - 2x + 1 = -\frac{1}{2} + 1$，
即$(x - 1)^2 = \frac{1}{2}$，
则$x - 1 = \frac{\sqrt{2}}{2}$，或$x - 1 = -\frac{\sqrt{2}}{2}$，
$\therefore x_1 = 1 + \frac{\sqrt{2}}{2}$，$x_2 = 1 - \frac{\sqrt{2}}{2}$.
(2)方程的两边同除以2，得$x^2 - \frac{5}{2}x + \frac{3}{2} = 0$.
移项，得$x^2 - \frac{5}{2}x = -\frac{3}{2}$.
方程的两边同加上$\frac{25}{16}$，得$x^2 - \frac{5}{2}x + \frac{25}{16} = -\frac{3}{2} + \frac{25}{16}$，
即$(x - \frac{5}{4})^2 = \frac{1}{16}$，
则$x - \frac{5}{4} = \frac{1}{4}$，或$x - \frac{5}{4} = -\frac{1}{4}$，
$\therefore x_1 = \frac{3}{2}$，$x_2 = 1$.
(3)方程的两边同乘4，得$x^2 - 24x + 12 = 0$.
移项，得$x^2 - 24x = -12$.
方程的两边同加上144，得$x^2 - 24x + 144 = -12 + 144$，
即$(x - 12)^2 = 132$，
则$x - 12 = 2\sqrt{33}$，或$x - 12 = -2\sqrt{33}$，
$\therefore x_1 = 12 + 2\sqrt{33}$，$x_2 = 12 - 2\sqrt{33}$.
(1)方程的两边同除以2，得$x^2 - 2x + \frac{1}{2} = 0$.
移项，得$x^2 - 2x = -\frac{1}{2}$.
方程的两边同加上1，得$x^2 - 2x + 1 = -\frac{1}{2} + 1$，
即$(x - 1)^2 = \frac{1}{2}$，
则$x - 1 = \frac{\sqrt{2}}{2}$，或$x - 1 = -\frac{\sqrt{2}}{2}$，
$\therefore x_1 = 1 + \frac{\sqrt{2}}{2}$，$x_2 = 1 - \frac{\sqrt{2}}{2}$.
(2)方程的两边同除以2，得$x^2 - \frac{5}{2}x + \frac{3}{2} = 0$.
移项，得$x^2 - \frac{5}{2}x = -\frac{3}{2}$.
方程的两边同加上$\frac{25}{16}$，得$x^2 - \frac{5}{2}x + \frac{25}{16} = -\frac{3}{2} + \frac{25}{16}$，
即$(x - \frac{5}{4})^2 = \frac{1}{16}$，
则$x - \frac{5}{4} = \frac{1}{4}$，或$x - \frac{5}{4} = -\frac{1}{4}$，
$\therefore x_1 = \frac{3}{2}$，$x_2 = 1$.
(3)方程的两边同乘4，得$x^2 - 24x + 12 = 0$.
移项，得$x^2 - 24x = -12$.
方程的两边同加上144，得$x^2 - 24x + 144 = -12 + 144$，
即$(x - 12)^2 = 132$，
则$x - 12 = 2\sqrt{33}$，或$x - 12 = -2\sqrt{33}$，
$\therefore x_1 = 12 + 2\sqrt{33}$，$x_2 = 12 - 2\sqrt{33}$.
7 已知A = 2x² + 7x - 1，B = 2 + x，A的值与B的值互为相反数，求x的值.
答案：
解：根据题意，得$2x^2 + 7x - 1 + 2 + x = 0$.
整理，得$2x^2 + 8x + 1 = 0$，
$\therefore x^2 + 4x + 4 = -\frac{1}{2} + 4$，
$\therefore (x + 2)^2 = \frac{7}{2}$，$\therefore x + 2 = \pm\frac{\sqrt{14}}{2}$，
$\therefore x_1 = -2 + \frac{\sqrt{14}}{2}$，$x_2 = -2 - \frac{\sqrt{14}}{2}$.
整理，得$2x^2 + 8x + 1 = 0$，
$\therefore x^2 + 4x + 4 = -\frac{1}{2} + 4$，
$\therefore (x + 2)^2 = \frac{7}{2}$，$\therefore x + 2 = \pm\frac{\sqrt{14}}{2}$，
$\therefore x_1 = -2 + \frac{\sqrt{14}}{2}$，$x_2 = -2 - \frac{\sqrt{14}}{2}$.
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