1. 将下列各式配方：
(1)$2x^2 + 5x + 4 = 2(x + \_\_\_\_\_\_)^2 + \_\_\_\_\_\_;$
(2)$\frac{1}{3}x^2 - x - 5 = \frac{1}{3}(x - \_\_\_\_\_\_)^2 + \_\_\_\_\_\_.$

(1)$\dfrac{5}{4}$ $\dfrac{7}{8}$;(2)$\dfrac{3}{2}$ $-\dfrac{23}{4}$

2. [2025南京秦淮区期中] 下列解方程$2x^2 -4x = -1$的步骤中,依据是“平方根的意义”的是（）

A.第一步:两边都除以2,得$x^2 -2x = -\dfrac{1}{2}$
B.第二步:配方,得$x^2 -2x +1 = -\dfrac{1}{2} +1$,即$(x-1)^2 = \dfrac{1}{2}$
C.第三步:开平方,得$x -1 = \pm\dfrac{\sqrt{2}}{2}$
D.第四步:移项,得$x = 1 \pm\dfrac{\sqrt{2}}{2}$,即$x_1 = 1 + \dfrac{\sqrt{2}}{2}$,$x_2 = 1 - \dfrac{\sqrt{2}}{2}$

C

3. [2026苏州工业园区期中]用配方法解一元二次方程$2x^2 - 2x -1 =0$，下列配方正确的是（）

A.$(x - \dfrac{1}{4})^2 = \dfrac{3}{4}$
B.$(x - \dfrac{1}{4})^2 = \dfrac{3}{2}$
C.$(x - \dfrac{1}{2})^2 = \dfrac{3}{4}$
D.$(x - \dfrac{1}{2})^2 = \dfrac{3}{2}$

C 两边都除以2,得$x^2 - x - \dfrac{1}{2} = 0$. 移项,得$x^2 - x = \dfrac{1}{2}$.配方,得$x^2 - x + (\dfrac{1}{2})^2 = \dfrac{1}{2} + (\dfrac{1}{2})^2$,即$(x - \dfrac{1}{2})^2 = \dfrac{3}{4}$.

4. ▶一题多解 [2026 南通如皋初级中学期中] 用配方法解一元二次方程$2x^2 - bx + a = 0$，得$x - \frac{3}{2} = \pm \frac{\sqrt{15}}{2}$，则$b$的值为。

6
快解 将方程$x - \dfrac{3}{2} = \pm \dfrac{\sqrt{15}}{2}$两边平方,得$(x - \dfrac{3}{2})^2 = \dfrac{15}{4}$,展开,得$x^2 - 3x + \dfrac{9}{4} = \dfrac{15}{4}$,整理得$x^2 - 3x - \dfrac{3}{2} = 0$,$\therefore 2x^2 - 6x - 3 = 0$,与原方程$2x^2 - bx + a = 0$比较系数,得$b=6$.
通解 $2x^2 - bx + a = 0$,移项,得$2x^2 - bx = -a$. 二次项系数化为1,得$x^2 - \dfrac{b}{2}x = -\dfrac{a}{2}$. 配方,得$x^2 - \dfrac{b}{2}x + (\dfrac{b}{4})^2 = -\dfrac{a}{2} + (\dfrac{b}{4})^2$,即$(x - \dfrac{b}{4})^2 = -\dfrac{a}{2} + \dfrac{b^2}{16}$. $\because$ 用配方法解一元二次方程$2x^2 - bx + a = 0$得$x - \dfrac{3}{2} = \pm \dfrac{\sqrt{15}}{2}$,$\therefore \dfrac{b}{4} = \dfrac{3}{2}$,$\therefore b=6$.

5 一题多解 将一元二次方程$2x^2 +12x =5$配方后得到$2(x+c)^2 =b$，则$b+c=$.

26
通解 $2x^2 +12x =5$,$2(x^2 +6x) =5$,$2(x^2 +6x +9 -9) =5$,$2(x+3)^2 -18 =5$,$2(x+3)^2 =5+18$,即$2(x+3)^2 =23$.
$\because$ 方程$2x^2 +12x =5$配方后得到$2(x+c)^2 =b$,$\therefore b=23$,$c=3$,$\therefore b+c=23+3=26$.
另解 $\because 2(x+c)^2 =b$,$\therefore 2(x^2 +2cx +c^2) =b$,即$2x^2 +4cx =b -2c^2$. $\because$ 方程$2x^2 +12x =5$配方后得到$2(x+c)^2 =b$,$\therefore 4c=12$,$b -2c^2 =5$,$\therefore c=3$,$b=23$,$\therefore b+c=23+3=26$.

6. 用配方法解下列方程:
(1)$3x^2 - 6x - 4 = 0$;
(2)$\frac{1}{2}x^2 - 8x + \frac{1}{2} = 0$;
(3)$4x^2 - 3 = -4x$;
(4)$-3x^2 + 6x - 1 = 0$.

解:(1)方程两边都除以3,得$x^2 - 2x - \dfrac{4}{3} = 0$.
移项,得$x^2 - 2x = \dfrac{4}{3}$.
配方,得$x^2 - 2x + 1 = \dfrac{4}{3} + 1$,即$(x-1)^2 = \dfrac{7}{3}$.
解这个方程,得$x - 1 = \pm \dfrac{\sqrt{21}}{3}$,
所以$x_1 = 1 + \dfrac{\sqrt{21}}{3}$,$x_2 = 1 - \dfrac{\sqrt{21}}{3}$.
(2)方程两边都乘2,得$x^2 - 16x + 1 = 0$.
移项,得$x^2 - 16x = -1$.
配方,得$x^2 - 16x + (-8)^2 = -1 + (-8)^2$,
即$(x-8)^2 = 63$.
解这个方程,得$x - 8 = \pm 3\sqrt{7}$,
所以$x_1 = 8 + 3\sqrt{7}$,$x_2 = 8 - 3\sqrt{7}$,
(3)方程两边都除以4,得$x^2 - \dfrac{3}{4} = -x$.
移项,得$x^2 + x = \dfrac{3}{4}$.
配方,得$x^2 + x + (\dfrac{1}{2})^2 = \dfrac{3}{4} + (\dfrac{1}{2})^2$,
即$(x + \dfrac{1}{2})^2 = 1$.
解这个方程,得$x + \dfrac{1}{2} = \pm 1$,
所以$x_1 = \dfrac{1}{2}$,$x_2 = -\dfrac{3}{2}$.
(4)方程两边都除以$-3$,得$x^2 - 2x + \dfrac{1}{3} = 0$.
移项,得$x^2 - 2x = -\dfrac{1}{3}$.
配方,得$x^2 - 2x + 1 = -\dfrac{1}{3} + 1$,
即$(x-1)^2 = \dfrac{2}{3}$.
解这个方程,得$x - 1 = \pm \dfrac{\sqrt{6}}{3}$,
所以$x_1 = 1 + \dfrac{\sqrt{6}}{3}$,$x_2 = 1 - \dfrac{\sqrt{6}}{3}$.

7. 大家知道在用配方法解一般形式的一元二次方程时,都要先把二次项系数化为1,再进行配方. 现请你阅读如下解方程①的过程,并按照此方法解方程②.
$2x^2 - 2\sqrt{2}x - 3 = 0$. ①
解:$2x^2 - 2\sqrt{2}x - 3 = 0$,
则$(\sqrt{2}x)^2 - 2\sqrt{2}x + 1 = 3 + 1$,
则$(\sqrt{2}x - 1)^2 = 4$,
直接开平方,得$\sqrt{2}x - 1 = \pm 2$,
所以 $x_1 = -\frac{\sqrt{2}}{2}, x_2 = \frac{3\sqrt{2}}{2}$.
$3x^2 - 2\sqrt{6}x = 2$. ②

解:$3x^2 - 2\sqrt{6}x = 2$,
则$(\sqrt{3}x)^2 - 2 × \sqrt{3}x × \sqrt{2} + (\sqrt{2})^2 = 2 + (\sqrt{2})^2$,
则$(\sqrt{3}x - \sqrt{2})^2 = 4$,
直接开平方,得$\sqrt{3}x - \sqrt{2} = \pm 2$,
所以$x_1 = \dfrac{\sqrt{6} + 2\sqrt{3}}{3}$,$x_2 = \dfrac{\sqrt{6} - 2\sqrt{3}}{3}$.