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1. 教材练习变式 填空：
(1) $x^2 + 6x + \_\_\_\_\_\_ = (x + \_\_\_\_\_\_)^2$；
(2) $x^2 - 5x + \_\_\_\_\_\_ = (x - \_\_\_\_\_\_)^2$；
(3) $x^2 - \frac{2}{5}x + \_\_\_\_\_\_ = (x - \_\_\_\_\_\_)^2$；
(4) $x^2 + 4\sqrt{2}x + \_\_\_\_\_\_ = (x + \_\_\_\_\_\_)^2$。
(1) $x^2 + 6x + \_\_\_\_\_\_ = (x + \_\_\_\_\_\_)^2$；
(2) $x^2 - 5x + \_\_\_\_\_\_ = (x - \_\_\_\_\_\_)^2$；
(3) $x^2 - \frac{2}{5}x + \_\_\_\_\_\_ = (x - \_\_\_\_\_\_)^2$；
(4) $x^2 + 4\sqrt{2}x + \_\_\_\_\_\_ = (x + \_\_\_\_\_\_)^2$。
答案
(1)$9\ \ 3$;(2)$\frac{25}{4}\ \ \frac{5}{2}$;(3)$\frac{1}{25}\ \ \frac{1}{5}$;(4)$8\ \ 2\sqrt{2}$
2. [2026无锡宜城中学月考] 关于x的一元二次方程$x^2 -4x -1=0$用配方法可变形为（
A.$(x+2)^2=3$
B.$(x+2)^2=5$
C.$(x-2)^2=5$
D.$(x-2)^2=3$
C
）A.$(x+2)^2=3$
B.$(x+2)^2=5$
C.$(x-2)^2=5$
D.$(x-2)^2=3$
答案
移项,得$x^2 -4x = 1$. 配方,得$x^2 - 2 · x · 2 + 2^2 = 1 + 2^2$,$(x - 2)^2 = 5$.
3. ▶一题多解 [2024东营中考] 用配方法解一元二次方程$x^2 -2x -2023=0$，将它转化为$(x+a)^2 = b$的形式，则$a^b$的值为 (
A.$-2024$
B.$2024$
C.$-1$
D.$1$
D
)A.$-2024$
B.$2024$
C.$-1$
D.$1$
答案
将$x^2 - 2x - 2\ 023 = 0$配方,得$x^2 - 2x + 1 = 2\ 024$,即$(x - 1)^2 = 2\ 024$,$\therefore a = - 1$,$b = 2\ 024$,$\therefore a^b = (-1)^{2\ 024} = 1$.
另解 由$(x + a)^2 = b$得$x^2 + 2ax + a^2 = b$,$\therefore 2a = - 2$,$a^2 - b = - 2\ 023$,$\therefore a = - 1$,$b = 2\ 024$,$\therefore a^b = (-1)^{2\ 024} = 1$.
另解 由$(x + a)^2 = b$得$x^2 + 2ax + a^2 = b$,$\therefore 2a = - 2$,$a^2 - b = - 2\ 023$,$\therefore a = - 1$,$b = 2\ 024$,$\therefore a^b = (-1)^{2\ 024} = 1$.
4. 若一元二次方程$x^2 -4x +b =0$可转化为$(x - a)^2 =3$的形式，则$a + b$的值为
3
。答案
由$x^2 -4x +b =0$,得$x^2 -4x = -b$,则$x^2 -4x +4 =4 -b$,即$(x - 2)^2 =4 -b$,它与$(x - a)^2 =3$相同,$\therefore a = 2$,$4 - b = 3$,$\therefore b = 1$,$\therefore a + b = 3$.
5. 教材阅读变式 如图,在用配方法解一元二次方程$x^2 + 6x = 40$时,配方的过程可以用拼图直观地表示,即看成将一个长是$(x+6)$、宽是$x$、面积是40的矩形割补成一个正方形,则$m$的值是
3
.答案
将$x^2 +6x =40$配方,得$x^2 +6x +9 =40 +9$,即$(x +3)^2 =49$,所以$m = 3$.
6. 解下列方程:
(1)$x^2 + 10x = 2$;
(2)$x^2 - 8x - 1 = 0$;
(3)$x^2 + 2 = -3x$;
(4)$x^2 - \frac{5}{2}x + \frac{1}{2} = 0$.
(1)$x^2 + 10x = 2$;
(2)$x^2 - 8x - 1 = 0$;
(3)$x^2 + 2 = -3x$;
(4)$x^2 - \frac{5}{2}x + \frac{1}{2} = 0$.
答案
(1)配方,得$x^2 + 2 · x · 5 + 5^2 = 2 + 5^2$,
即$(x + 5)^2 = 27$.
解这个方程,得$x + 5 = \pm 3\sqrt{3}$,
所以$x_1 = 3\sqrt{3} - 5$,$x_2 = - 3\sqrt{3} - 5$.
(2)移项,得$x^2 - 8x = 1$.
配方,得$x^2 - 8x + 4^2 = 1 + 4^2$,
即$(x - 4)^2 = 17$.
解这个方程,得$x - 4 = \pm \sqrt{17}$,
所以$x_1 = 4 + \sqrt{17}$,$x_2 = 4 - \sqrt{17}$.
(3)移项,得$x^2 + 3x = - 2$.
配方,得$x^2 + 3x + (\frac{3}{2})^2 = - 2 + (\frac{3}{2})^2$,
即$(x + \frac{3}{2})^2 = \frac{1}{4}$.
解这个方程,得$x + \frac{3}{2} = \pm \frac{1}{2}$,
所以$x_1 = - 1$,$x_2 = - 2$.
(4)移项,得$x^2 - \frac{5}{2}x = - \frac{1}{2}$.
配方,得$x^2 - 2 · x · \frac{5}{4} + (\frac{5}{4})^2 = - \frac{1}{2} + (\frac{5}{4})^2$,
即$(x - \frac{5}{4})^2 = \frac{17}{16}$.
解这个方程,得$x - \frac{5}{4} = \pm \frac{\sqrt{17}}{4}$,
所以$x_1 = \frac{\sqrt{17}}{4} + \frac{5}{4}$,$x_2 = - \frac{\sqrt{17}}{4} + \frac{5}{4}$.
即$(x + 5)^2 = 27$.
解这个方程,得$x + 5 = \pm 3\sqrt{3}$,
所以$x_1 = 3\sqrt{3} - 5$,$x_2 = - 3\sqrt{3} - 5$.
(2)移项,得$x^2 - 8x = 1$.
配方,得$x^2 - 8x + 4^2 = 1 + 4^2$,
即$(x - 4)^2 = 17$.
解这个方程,得$x - 4 = \pm \sqrt{17}$,
所以$x_1 = 4 + \sqrt{17}$,$x_2 = 4 - \sqrt{17}$.
(3)移项,得$x^2 + 3x = - 2$.
配方,得$x^2 + 3x + (\frac{3}{2})^2 = - 2 + (\frac{3}{2})^2$,
即$(x + \frac{3}{2})^2 = \frac{1}{4}$.
解这个方程,得$x + \frac{3}{2} = \pm \frac{1}{2}$,
所以$x_1 = - 1$,$x_2 = - 2$.
(4)移项,得$x^2 - \frac{5}{2}x = - \frac{1}{2}$.
配方,得$x^2 - 2 · x · \frac{5}{4} + (\frac{5}{4})^2 = - \frac{1}{2} + (\frac{5}{4})^2$,
即$(x - \frac{5}{4})^2 = \frac{17}{16}$.
解这个方程,得$x - \frac{5}{4} = \pm \frac{\sqrt{17}}{4}$,
所以$x_1 = \frac{\sqrt{17}}{4} + \frac{5}{4}$,$x_2 = - \frac{\sqrt{17}}{4} + \frac{5}{4}$.
7.若一元二次方程$x^2 -2x -99=0$的两根分别为$a,b$,且$a ＞ b$,求$2a -b$的值.
答案
将方程$x^2 -2x -99 = 0$移项,得$x^2 - 2x = 99$,
配方,得$x^2 - 2x + 1 = 99 + 1$,
整理,得$(x - 1)^2 = 100$,
解这个方程,得$x - 1 = \pm 10$,
所以$x_1 = 11$,$x_2 = - 9$.
由题意,得$a = 11$,$b = - 9$,
所以$2a - b = 2 × 11 - (-9) = 31$.
配方,得$x^2 - 2x + 1 = 99 + 1$,
整理,得$(x - 1)^2 = 100$,
解这个方程,得$x - 1 = \pm 10$,
所以$x_1 = 11$,$x_2 = - 9$.
由题意,得$a = 11$,$b = - 9$,
所以$2a - b = 2 × 11 - (-9) = 31$.
8. 小静同学解方程$x^2 + 2x -8=0$的步骤为：①移项，得$x^2 +2x=8$；②配方，得$x^2 +2x +1=8+1$；③即$(x+1)^2=9$；④解这个方程，得$x+1=\pm3$；⑤即$x=1\pm3$；⑥所以$x_1=4,x_2=-2$。
(1)小静是从步骤
(2)用配方法解方程$x^2 +2nx -8n^2=0$.(用含有$n$的式子表示方程的根)
(1)小静是从步骤
⑤
开始出现错误的。(2)用配方法解方程$x^2 +2nx -8n^2=0$.(用含有$n$的式子表示方程的根)
答案
(1)⑤
(2)移项,得$x^2 + 2nx = 8n^2$.
配方,得$x^2 + 2nx + n^2 = 8n^2 + n^2$,即$(x + n)^2 = 9n^2$.
解这个方程,得$x + n = \pm 3n$,
所以$x_1 = 2n$,$x_2 = - 4n$.
(2)移项,得$x^2 + 2nx = 8n^2$.
配方,得$x^2 + 2nx + n^2 = 8n^2 + n^2$,即$(x + n)^2 = 9n^2$.
解这个方程,得$x + n = \pm 3n$,
所以$x_1 = 2n$,$x_2 = - 4n$.
