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1. 用配方法解方程$x^2 +6x=1$时，方程的左、右两边需同时加上 （
A.3
B.6
C.9
D.36
C
）A.3
B.6
C.9
D.36
答案
C
2. 用配方法解一元二次方程$x^2 - 2x - 2026 = 0$，将它转化为$(x+a)^2 = b$的形式，则$a^b$的值为（
A.2026
B.2027
C.$-1$
D.1
C
）A.2026
B.2027
C.$-1$
D.1
答案
C 【点拨】$\because x^2 - 2x - 2\ 026 = 0, \therefore x^2 - 2x = 2\ 026.$
$\therefore x^2 - 2x + 1 = 2\ 026 + 1. \therefore (x-1)^2 = 2\ 027. \therefore a = -1,$
$b = 2\ 027. \therefore a^b = (-1)^{2\ 027} = -1.$
$\therefore x^2 - 2x + 1 = 2\ 026 + 1. \therefore (x-1)^2 = 2\ 027. \therefore a = -1,$
$b = 2\ 027. \therefore a^b = (-1)^{2\ 027} = -1.$
3. 母题 教材P₂₇练习T₁ 填空：
(1)$x^2 + 10x +$
(2)$x^2 - 12x +$
(3)$x^2 + 5x +$
(4)$x^2 - \frac{2}{3}x +$
(1)$x^2 + 10x +$
25
$=(x +$ 5
$)^2$；(2)$x^2 - 12x +$
36
$=(x -$ 6
$)^2$；(3)$x^2 + 5x +$
$\dfrac{25}{4}$
$=(x +$ $\dfrac{5}{2}$
$)^2$；(4)$x^2 - \frac{2}{3}x +$
$\dfrac{1}{9}$
$=(x -$ $\dfrac{1}{3}$
$)^2$。答案
3. (1)25;5 (2)36;6 (3)$\dfrac{25}{4};\dfrac{5}{2}$ (4)$\dfrac{1}{9};\dfrac{1}{3}$
4. 用配方法解方程：
(1)$x^2 - 2x = 3$；
(2)$x^2 - 2\sqrt{2}x + 1 = 0$。
(1)$x^2 - 2x = 3$；
(2)$x^2 - 2\sqrt{2}x + 1 = 0$。
答案
【解】(1)$x^2 - 2x = 3, (x-1)^2 = 4, x-1 = \pm 2, x_1 = 3,$
$x_2 = -1.$
(2)移项,得$x^2 - 2\sqrt{2}\ x = -1.$ 配方,得$x^2 - 2\sqrt{2}\ x + 2 = -1 + 2,$即$(x-\sqrt{2})^2 = 1,$解得$x_1 = \sqrt{2} + 1, x_2 = \sqrt{2} - 1.$
$x_2 = -1.$
(2)移项,得$x^2 - 2\sqrt{2}\ x = -1.$ 配方,得$x^2 - 2\sqrt{2}\ x + 2 = -1 + 2,$即$(x-\sqrt{2})^2 = 1,$解得$x_1 = \sqrt{2} + 1, x_2 = \sqrt{2} - 1.$
5. 用配方法解方程$2x^2 - 3x -1=0$，配方正确的是（
A.$(x-\dfrac{3}{4})^2=\dfrac{17}{16}$
B.$(x-\dfrac{3}{4})^2=\dfrac{1}{2}$
C.$(x-\dfrac{3}{2})^2=\dfrac{13}{4}$
D.$(x-\dfrac{3}{2})^2=\dfrac{11}{4}$
A
）A.$(x-\dfrac{3}{4})^2=\dfrac{17}{16}$
B.$(x-\dfrac{3}{4})^2=\dfrac{1}{2}$
C.$(x-\dfrac{3}{2})^2=\dfrac{13}{4}$
D.$(x-\dfrac{3}{2})^2=\dfrac{11}{4}$
答案
A
6. 新考法 过程辨析法 某数学兴趣小组的四人以接龙的方式用配方法解一元二次方程，每人只负责完成一个步骤(如图)，老师看后，发现最后结果是错误的，并说：“错误是从某名同学负责的步骤开始出现的.”则这名同学是 (
A.甲
B.乙
C.丙
D.丁
B
)A.甲
B.乙
C.丙
D.丁
答案
B
7. 用配方法解方程：
(1)$\frac{1}{2}x^2 +4x -6=0$；
(2)$(x+1)(2x-3)=1$。
(1)$\frac{1}{2}x^2 +4x -6=0$；
(2)$(x+1)(2x-3)=1$。
答案
【解】(1)原方程可化为$x^2 + 8x = 12,$配方,得$x^2 + 8x + 4^2 = 12 + 4^2,$
即$(x+4)^2 = 28, \therefore x+4 = \pm 2\sqrt{7}.$
$\therefore x_1 = 2\sqrt{7} - 4, x_2 = -2\sqrt{7} - 4.$
(2)去括号、移项、合并同类项,得$2x^2 - x = 4.$ 二次项系数化为1,得$x^2 - \dfrac{1}{2}x = 2.$ 配方,得$x^2 - \dfrac{1}{2}x + \dfrac{1}{16} = 2 + \dfrac{1}{16},$即$(x - \dfrac{1}{4})^2 = \dfrac{33}{16}, \therefore x - \dfrac{1}{4} = \pm \dfrac{\sqrt{33}}{4}, \therefore x_1 = \dfrac{1+\sqrt{33}}{4}, x_2 = \dfrac{1-\sqrt{33}}{4}.$
即$(x+4)^2 = 28, \therefore x+4 = \pm 2\sqrt{7}.$
$\therefore x_1 = 2\sqrt{7} - 4, x_2 = -2\sqrt{7} - 4.$
(2)去括号、移项、合并同类项,得$2x^2 - x = 4.$ 二次项系数化为1,得$x^2 - \dfrac{1}{2}x = 2.$ 配方,得$x^2 - \dfrac{1}{2}x + \dfrac{1}{16} = 2 + \dfrac{1}{16},$即$(x - \dfrac{1}{4})^2 = \dfrac{33}{16}, \therefore x - \dfrac{1}{4} = \pm \dfrac{\sqrt{33}}{4}, \therefore x_1 = \dfrac{1+\sqrt{33}}{4}, x_2 = \dfrac{1-\sqrt{33}}{4}.$
8. 新考法作差比较法 已知$P=\frac{7}{15}m -1$，$Q=m^2 - \frac{8}{15}m$（m为任意实数），则P，Q的大小关系为（
A.$P＞Q$
B.$P＜Q$
C.$P≥ Q$
D.$P≤ Q$
B
）A.$P＞Q$
B.$P＜Q$
C.$P≥ Q$
D.$P≤ Q$
答案
B 【点拨】由题意,知$Q-P = m^2 - \dfrac{8}{15}m - \dfrac{7}{15}m + 1 = m^2 - m + 1 = m^2 - m + \dfrac{1}{4} + \dfrac{3}{4} = (m - \dfrac{1}{2})^2 + \dfrac{3}{4}.$
$\because (m - \dfrac{1}{2})^2 \ge 0, \therefore (m - \dfrac{1}{2})^2 + \dfrac{3}{4} ＞ 0, \therefore Q-P ＞ 0,$即$P＜Q.$
$\because (m - \dfrac{1}{2})^2 \ge 0, \therefore (m - \dfrac{1}{2})^2 + \dfrac{3}{4} ＞ 0, \therefore Q-P ＞ 0,$即$P＜Q.$
9.方程$x^2 -2x -1=0$的较小的根为$m$，方程$x^2 -2\sqrt{2}x -2=0$的较大的根为$n$，则$m+n$等于（
A.$3$
B.$-3$
C.$2\sqrt{2}$
D.$-2\sqrt{2}$
A
）A.$3$
B.$-3$
C.$2\sqrt{2}$
D.$-2\sqrt{2}$
答案
A 【点拨】由$x^2 - 2x - 1 = 0$得$x = 1\pm\sqrt{2}.$
$\because m$是较小的根，
$\therefore m=1-\sqrt{2}.$ 由$x^2 - 2\sqrt{2}x - 2 = 0$得$x = \sqrt{2}\pm2.$
$\because n$是较大的根，$\therefore n=\sqrt{2}+2,$
$\therefore m+n=1-\sqrt{2}+\sqrt{2}+2=3.$
$\because m$是较小的根，
$\therefore m=1-\sqrt{2}.$ 由$x^2 - 2\sqrt{2}x - 2 = 0$得$x = \sqrt{2}\pm2.$
$\because n$是较大的根，$\therefore n=\sqrt{2}+2,$
$\therefore m+n=1-\sqrt{2}+\sqrt{2}+2=3.$
10. 新考向数学文化 欧几里得的《几何原本》中记载了形如$x^2 - 2bx + 4c^2 = 0(b＞2c＞0)$的方程根的图形解法:如图,画$Rt△ ABC$,使$∠ ACB=90°,AC=2c,AB=b$,以B为圆心,BC长为半径画圆,交射线AB于点D,E,则该方程较大的根是 (
A.CE 的长度
B.CD 的长度
C.DE 的长度
D.AE 的长度
D
)A.CE 的长度
B.CD 的长度
C.DE 的长度
D.AE 的长度
答案
D 【点拨】$\because x^2 - 2bx + 4c^2 = 0, \therefore x^2 - 2bx = -4c^2.$
$\therefore x^2 - 2bx + b^2 = b^2 - 4c^2. \therefore (x-b)^2 = b^2 - 4c^2. \therefore x-b = \pm \sqrt{b^2 - 4c^2}.$
$\therefore x_1 = b + \sqrt{b^2 - 4c^2}, x_2 = b - \sqrt{b^2 - 4c^2}.$
$\because$在Rt$△ ABC$中,$∠ ACB=90°, AC=2c, AB=b, \therefore BC=\sqrt{AB^2 - AC^2}=\sqrt{b^2 - 4c^2}, \therefore$方程较大的根是$AB+BC=AB+BE=AE$的长度,故选D.
$\therefore x^2 - 2bx + b^2 = b^2 - 4c^2. \therefore (x-b)^2 = b^2 - 4c^2. \therefore x-b = \pm \sqrt{b^2 - 4c^2}.$
$\therefore x_1 = b + \sqrt{b^2 - 4c^2}, x_2 = b - \sqrt{b^2 - 4c^2}.$
$\because$在Rt$△ ABC$中,$∠ ACB=90°, AC=2c, AB=b, \therefore BC=\sqrt{AB^2 - AC^2}=\sqrt{b^2 - 4c^2}, \therefore$方程较大的根是$AB+BC=AB+BE=AE$的长度,故选D.
