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1.应用完全平方公式计算$(m-2n+p)^2$时，下列变形错误的是（
A.$[(m-2n)+p]^2$
B.$[m-(2n-p)]^2$
C.$[(m+p)-2n]^2$
D.$[m-(2n+p)]^2$
D
）A.$[(m-2n)+p]^2$
B.$[m-(2n-p)]^2$
C.$[(m+p)-2n]^2$
D.$[m-(2n+p)]^2$
答案
1.D
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2.「2026广东湛江月考」下列添括号正确的是 (
A.$b+c=-(b+c)$
B.$-2x+4y=-2(x-4y)$
C.$a-b=+(a-b)$
D.$2x-y-1=2x-(y-1)$
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2.「2026广东湛江月考」下列添括号正确的是 (
C
)A.$b+c=-(b+c)$
B.$-2x+4y=-2(x-4y)$
C.$a-b=+(a-b)$
D.$2x-y-1=2x-(y-1)$
答案
2.C b+c=+(b+c)=-(-b-c),-2x+4y=-2(x-2y),a-b=+(a-b),2x-y-1=2x-(y+1),只有C选项正确.故选C.
3.在括号内填上适当的项.
(1)$a - 2b + c + d = a - ($
(2)$-a - 3b + c = -($
(3)$x^2 - 2y^2 + 2x - 3y = ($
(4)$x^2 - y^2 - x - y = x^2 - x - ($
(1)$a - 2b + c + d = a - ($
2b-c-d
$)$.(2)$-a - 3b + c = -($
a+3b-c
$)$.(3)$x^2 - 2y^2 + 2x - 3y = ($
x²-2y²
$) + 2x - 3y$.(4)$x^2 - y^2 - x - y = x^2 - x - ($
y²+y
$)$.答案
3.(1)2b-c-d (2)a+3b-c (3)$x^2-2y^2$ (4)$y^2+y$
4. 学科 特色 教材变式 计算：
(1) $(a+b-1)^2$.
(2) $(x-2y-1)^2$.
(3) $(a^2 - a + 1)(a^2 + a + 1)$.
(4) $(3m + 2n - p)(3m - 2n + p)$.
(1) $(a+b-1)^2$.
(2) $(x-2y-1)^2$.
(3) $(a^2 - a + 1)(a^2 + a + 1)$.
(4) $(3m + 2n - p)(3m - 2n + p)$.
答案
4.解析
(1)$(a+b-1)^2=(a+b)^2-2(a+b)+1$
=$a^2+2ab+b^2-2a-2b+1.$
(2)$(x-2y-1)^2=(x-2y)^2-2(x-2y)+1$
=$x^2-4xy+4y^2-2x+4y+1.$
(3)$(a^2-a+1)(a^2+a+1)$
=$[(a^2+1)-a][(a^2+1)+a]$
=$(a^2+1)^2-a^2=a^4+2a^2+1-a^2=a^4+a^2+1.$
(4)$(3m+2n-p)(3m-2n+p)$
=$[3m+(2n-p)][3m-(2n-p)]$
=$(3m)^2-(2n-p)^2=9m^2-4n^2+4np-p^2.$
(1)$(a+b-1)^2=(a+b)^2-2(a+b)+1$
=$a^2+2ab+b^2-2a-2b+1.$
(2)$(x-2y-1)^2=(x-2y)^2-2(x-2y)+1$
=$x^2-4xy+4y^2-2x+4y+1.$
(3)$(a^2-a+1)(a^2+a+1)$
=$[(a^2+1)-a][(a^2+1)+a]$
=$(a^2+1)^2-a^2=a^4+2a^2+1-a^2=a^4+a^2+1.$
(4)$(3m+2n-p)(3m-2n+p)$
=$[3m+(2n-p)][3m-(2n-p)]$
=$(3m)^2-(2n-p)^2=9m^2-4n^2+4np-p^2.$
5.「2026广东广州期中」先化简,再求值:
$(x+y+2)(x+y-2)-(x+2y)^2+3y^2$,其中$x=-1,y=3$.
$(x+y+2)(x+y-2)-(x+2y)^2+3y^2$,其中$x=-1,y=3$.
答案
5.解析 原式=$(x+y)^2-4-(x^2+4xy+4y^2)+3y^2$
=$x^2+2xy+y^2-4-x^2-4xy-4y^2+3y^2=-2xy-4$,
当$x=-1,y=3$时,
原式=$-2×(-1)×3-4=6-4=2.$
=$x^2+2xy+y^2-4-x^2-4xy-4y^2+3y^2=-2xy-4$,
当$x=-1,y=3$时,
原式=$-2×(-1)×3-4=6-4=2.$
6.「2026山东德州月考，★☆」若$(a+b+c)^2=4028,(a-b+c)^2=24$，则$a^2+b^2+c^2+2ac$的值为（
A.2 024
B.2 025
C.2 026
D.4 050
C
）A.2 024
B.2 025
C.2 026
D.4 050
答案
6.C $\because (a+b+c)^2=(a+b)^2+2(a+b)c+c^2$
=$a^2+2ab+b^2+2ac+2bc+c^2$
=$a^2+b^2+c^2+2ab+2ac+2bc$,
$(a-b+c)^2=(a-b)^2+2(a-b)c+c^2$
=$a^2-2ab+b^2+2ac-2bc+c^2$
=$a^2+b^2+c^2-2ab+2ac-2bc$,
$\therefore (a+b+c)^2+(a-b+c)^2=2(a^2+b^2+c^2+2ac)$,
$\because (a+b+c)^2=4028,(a-b+c)^2=24$,
$\therefore 2(a^2+b^2+c^2+2ac)=4028+24=4052$,
$\therefore a^2+b^2+c^2+2ac=2026.$
=$a^2+2ab+b^2+2ac+2bc+c^2$
=$a^2+b^2+c^2+2ab+2ac+2bc$,
$(a-b+c)^2=(a-b)^2+2(a-b)c+c^2$
=$a^2-2ab+b^2+2ac-2bc+c^2$
=$a^2+b^2+c^2-2ab+2ac-2bc$,
$\therefore (a+b+c)^2+(a-b+c)^2=2(a^2+b^2+c^2+2ac)$,
$\because (a+b+c)^2=4028,(a-b+c)^2=24$,
$\therefore 2(a^2+b^2+c^2+2ac)=4028+24=4052$,
$\therefore a^2+b^2+c^2+2ac=2026.$
7.「2025河北廊坊月考，★☆」当$x=-1$时，$ax+b+1$的值为$-3$，求$(a-b-1)(1-a+b)$的值.
答案
7.解析 $\because$ 当x=-1时,ax+b+1的值为-3,
$\therefore -a+b+1=-3,\therefore -a+b=-4,\therefore a-b=4$,
$\therefore (a-b-1)(1-a+b)$
=$[(a-b)-1][1-(a-b)]=(4-1)×(1-4)=-9.$
$\therefore -a+b+1=-3,\therefore -a+b=-4,\therefore a-b=4$,
$\therefore (a-b-1)(1-a+b)$
=$[(a-b)-1][1-(a-b)]=(4-1)×(1-4)=-9.$
8.「★☆」若 $ m ≠ 0 $，$ Q=(m^2 - m + 1)(m^2 + m + 1) $，$ P=(m+1)^2(m-1)^2 $，猜想 $ Q $ 与 $ P $ 的大小关系，并证明你的猜想。
答案
8.解析 Q＞P.证明如下：
$Q-P=(m^2-m+1)(m^2+m+1)-(m+1)^2(m-1)^2$
=$(m^2+1)^2-m^2-(m^2-1)^2$
=$(m^2+1+m^2-1)(m^2+1-m^2+1)-m^2$
=$4m^2-m^2=3m^2.$
$\because m≠0,\therefore 3m^2＞0.\therefore Q-P＞0.\therefore Q＞P.$
$Q-P=(m^2-m+1)(m^2+m+1)-(m+1)^2(m-1)^2$
=$(m^2+1)^2-m^2-(m^2-1)^2$
=$(m^2+1+m^2-1)(m^2+1-m^2+1)-m^2$
=$4m^2-m^2=3m^2.$
$\because m≠0,\therefore 3m^2＞0.\therefore Q-P＞0.\therefore Q＞P.$
9.「2026浙江宁波月考，★☆」已知$A=3x+2y-9$，$B=3x-2y-9$，求当$A·B+4y^2$的值最小时$x$的值.
答案
9.解析 $A· B+4y^2=(3x+2y-9)(3x-2y-9)+4y^2$
=$[(3x-9)+2y][(3x-9)-2y]+4y^2$
=$(3x-9)^2-(2y)^2+4y^2$
=$(3x-9)^2-4y^2+4y^2=(3x-9)^2$,
$\because (3x-9)^2≥0$,
$\therefore A· B+4y^2$的最小值为0,此时3x-9=0,解得x=3,
$\therefore$ 当$A· B+4y^2$的值最小时,x的值为3.
=$[(3x-9)+2y][(3x-9)-2y]+4y^2$
=$(3x-9)^2-(2y)^2+4y^2$
=$(3x-9)^2-4y^2+4y^2=(3x-9)^2$,
$\because (3x-9)^2≥0$,
$\therefore A· B+4y^2$的最小值为0,此时3x-9=0,解得x=3,
$\therefore$ 当$A· B+4y^2$的值最小时,x的值为3.
