1. (2025·上海校级月考)若$(2a + b)^2 = (2a - b)^2 + ($
A.$4ab$
B.$-4ab$
C.$8ab$
D.$-8ab$
C
$)$成立,则括号内的式子是 (C
)A.$4ab$
B.$-4ab$
C.$8ab$
D.$-8ab$
答案：1. C 解析:$(2a + b)^2 - (2a - b)^2 = 8ab$.故选 C.
2. 计算$(1 + 3x)(3x - 1) + 9(\frac{1}{3} - x)(x + \frac{1}{3})$的结果是 (
A.$18x^2 - 2$
B.$2 - 18x^2$
C.$0$
D.$8x^2$
C
)A.$18x^2 - 2$
B.$2 - 18x^2$
C.$0$
D.$8x^2$
答案：2. C 解析:$(1 + 3x)(3x - 1) = (3x + 1)(3x - 1) = (3x)^2 - 1^2 = 9x^2 - 1$,$9(\frac{1}{3} - x)(x + \frac{1}{3}) = 9[(\frac{1}{3})^2 - x^2] = 9(\frac{1}{9} - x^2) = 1 - 9x^2$,$(9x^2 - 1) + (1 - 9x^2) = 9x^2 - 9x^2 - 1 + 1 = 0$.故选 C.
3. (1)若$x + y = 3$,则$(x - y)^2 + 4xy + 1$的值为
(2)已知$(m - n)^2 = 36,(m + n)^2 = 400$,则$m^2 + n^2$的值为
10
;(2)已知$(m - n)^2 = 36,(m + n)^2 = 400$,则$m^2 + n^2$的值为
218
.答案：3. (1) 10 解析:因为$x + y = 3$,所以$x^2 + 2xy + y^2 = 9$,所以$(x - y)^2 + 4xy + 1 = x^2 + y^2 + 2xy + 1 = 10$.
(2) 218 解析:$(m - n)^2 + (m + n)^2 = 2(m^2 + n^2) = 436$,则$m^2 + n^2 = 218$.
(2) 218 解析:$(m - n)^2 + (m + n)^2 = 2(m^2 + n^2) = 436$,则$m^2 + n^2 = 218$.
4. (1)已知多项式$x^2 - 6x + m^2$是另一个多项式的平方,则$m =$
(2)若$4x^2 - mxy + \frac{1}{4}y^2$是一个完全平方式展开后得到的结果,则$m =$
$\pm 3$
;(2)若$4x^2 - mxy + \frac{1}{4}y^2$是一个完全平方式展开后得到的结果,则$m =$
$\pm 2$
.答案：4. (1) $\pm 3$ 解析:因为$x^2 - 6x + m^2$是一个完全平方式,所以$x^2 - 6x + m^2 = (x - 3)^2$,所以$m^2 = 9$,即$m = \pm 3$.
(2) $\pm 2$ 解析:$4x^2 - mxy + \frac{1}{4}y^2 = (2x)^2 - mxy + (\frac{1}{2}y)^2 = (2x \pm \frac{1}{2}y)^2$,所以$-mxy = \pm 2 · 2x · \frac{1}{2}y$,所以$m = \pm 2$.
(2) $\pm 2$ 解析:$4x^2 - mxy + \frac{1}{4}y^2 = (2x)^2 - mxy + (\frac{1}{2}y)^2 = (2x \pm \frac{1}{2}y)^2$,所以$-mxy = \pm 2 · 2x · \frac{1}{2}y$,所以$m = \pm 2$.
5. 计算:
(1)$(2m + 3n)^2 - (2m + n)(2m - n)$;
(2)$(m + 2n)^2(m - 2n)^2$;
(3)$(x - 3y - 2)^2$;
(4)$(a - 2b + 3c)(a + 2b - 3c)$.
(1)$(2m + 3n)^2 - (2m + n)(2m - n)$;
(2)$(m + 2n)^2(m - 2n)^2$;
(3)$(x - 3y - 2)^2$;
(4)$(a - 2b + 3c)(a + 2b - 3c)$.
答案：5. (1) 原式$= 4m^2 + 12mn + 9n^2 - (4m^2 - n^2) = 4m^2 + 12mn + 9n^2 - 4m^2 + n^2 = 10n^2 + 12mn$.
(2) 原式$= [(m + 2n)(m - 2n)]^2 = (m^2 - 4n^2)^2 = m^4 - 8m^2n^2 + 16n^4$.
(3) 原式$= [x - (3y + 2)]^2 = x^2 - 2x(3y + 2) + (3y + 2)^2 = x^2 + 9y^2 - 6xy - 4x + 12y + 4$.
(4) 原式$= [a - (2b - 3c)][a + (2b - 3c)] = a^2 - (2b - 3c)^2 = a^2 - (4b^2 - 12bc + 9c^2) = a^2 - 4b^2 + 12bc - 9c^2$.
(2) 原式$= [(m + 2n)(m - 2n)]^2 = (m^2 - 4n^2)^2 = m^4 - 8m^2n^2 + 16n^4$.
(3) 原式$= [x - (3y + 2)]^2 = x^2 - 2x(3y + 2) + (3y + 2)^2 = x^2 + 9y^2 - 6xy - 4x + 12y + 4$.
(4) 原式$= [a - (2b - 3c)][a + (2b - 3c)] = a^2 - (2b - 3c)^2 = a^2 - (4b^2 - 12bc + 9c^2) = a^2 - 4b^2 + 12bc - 9c^2$.
6. 先化简,再求值:
(1)$(x + 4)(x - 4) + (x - 3)^2$,其中$x = 3$;
(2)$(-x - 2y)(2y - x) - (2x - y)^2 + (3x - y)(2x - 5y)$,其中$x = -1,y = -2$.
(1)$(x + 4)(x - 4) + (x - 3)^2$,其中$x = 3$;
(2)$(-x - 2y)(2y - x) - (2x - y)^2 + (3x - y)(2x - 5y)$,其中$x = -1,y = -2$.
答案：6. (1) 原式$= x^2 - 16 + x^2 - 6x + 9 = 2x^2 - 6x - 7$.因为$x = 3$,所以原式$= 2 × 3^2 - 6 × 3 - 7 = -7$.
(2) 原式$= x^2 - 4y^2 - 4x^2 + 4xy - y^2 + 6x^2 - 15xy - 2xy + 5y^2 = 3x^2 - 13xy$,将$x = -1$,$y = -2$代入,得$3x^2 - 13xy = 3 × (-1)^2 - 13 × (-1) × (-2) = -23$.
(2) 原式$= x^2 - 4y^2 - 4x^2 + 4xy - y^2 + 6x^2 - 15xy - 2xy + 5y^2 = 3x^2 - 13xy$,将$x = -1$,$y = -2$代入,得$3x^2 - 13xy = 3 × (-1)^2 - 13 × (-1) × (-2) = -23$.
7. 如图,阴影部分是在边长为$a$的大正方形中剪去一个边长为$b$的小正方形后所得到的图形,将阴影部分通过割、拼,形成新的图形.给出下列2种割拼方法,其中能够验证平方差公式的是 (
A.①
B.②
C.①②
D.①②都不能
C
)A.①
B.②
C.①②
D.①②都不能
答案：7. C 解析:如题图①,左图阴影部分的面积为$a^2 - b^2$,右图的阴影部分是上底为$2b$,下底为$2a$,高为$(a - b)$的梯形,因此面积为$\frac{1}{2}(2b + 2a)(a - b) = (a + b)(a - b)$,所以有$a^2 - b^2 = (a + b)(a - b)$,因此题图①的割拼方法可以验证平方差公式.如题图②,左图阴影部分的面积为$a^2 - b^2$,右图的阴影部分是底为$(a + b)$,高为$(a - b)$的平行四边形,因此面积为$(a + b) · (a - b)$,所以有$a^2 - b^2 = (a + b)(a - b)$,因此题图②的割拼方法也可以验证平方差公式.故选 C.
8. 设$M = 2025^2 - 2022×2026,N = 2045^2 - 2045×3960 + 1980^2$,则$M$与$N$的大小关系为 (
A.$M ＞ N$
B.$M = N$
C.$M ＜ N$
D.无法确定
C
)A.$M ＞ N$
B.$M = N$
C.$M ＜ N$
D.无法确定
答案：8. C 解析:$M = 2025^2 - 2022 × 2026 = 2025^2 - (2024 - 2) × (2024 + 2) = 2025^2 - 2024^2 + 4 = (2025 + 2024)(2025 - 2024) + 4 = 4049 + 4 = 4053$.$N = 2045^2 - 2045 × 3960 + 1980^2 = 2045^2 - 2 × 2045 × 1980 + 1980^2 = (2045 - 1980)^2 = 65^2 = 4225$,所以$M ＜ N$.故选 C.