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1. 分解因式:$4a^2 -1 = \begin{pmatrix} \end{pmatrix}$
A.$(2a -1)(2a +1)$
B.$(a -2)(a +2)$
C.$(a -4)(a +1)$
D.$(4a -1)(a +1)$
A.$(2a -1)(2a +1)$
B.$(a -2)(a +2)$
C.$(a -4)(a +1)$
D.$(4a -1)(a +1)$
答案
A $4a^2 -1 = (2a)^2 -1 = (2a +1)(2a -1)$
2. [2025张掖期末] 下列多项式中，能用平方差公式分解因式的是 (
A.$a^2 + (-b)^2$
B.$5m^2 - 20mn$
C.$-x^2 - y^2$
D.$-x^2 + 9$
D
)A.$a^2 + (-b)^2$
B.$5m^2 - 20mn$
C.$-x^2 - y^2$
D.$-x^2 + 9$
答案
D $\because a^2 + (-b)^2 = a^2 + b^2 ,5m^2 -20mn = 5m(m -4n)$,$-x^2 - y^2 = -(x^2 + y^2)$,$\therefore$ 选项 A,B,C 都不能用平方差公式分解因式.$\because -x^2 +9 = -x^2 +3^2 = (3 +x)(3 -x)$,$\therefore$ 选项 D能用平方差公式分解因式.
3. [2025泰安期末]下列各式中，不能用两数和（差）的平方公式分解因式的是（
A.$x^2 - x + 0.25$
B.$16a^2 + 4a + 1$
C.$a^2 + 4ab + 4b^2$
D.$a^2 - 2a + 1$
B
）A.$x^2 - x + 0.25$
B.$16a^2 + 4a + 1$
C.$a^2 + 4ab + 4b^2$
D.$a^2 - 2a + 1$
答案
B $x^2 -x +0.25 = (x -0.5)^2$,$a^2 +4ab +4b^2 = (a +2b)^2$,$a^2 -2a +1 = (a -1)^2$.
4.已知$a^2 - b^2 = 12$，且$a - b = -2$，则$a + b =$
-6
。答案
-6 $\because a^2 - b^2 =12$,$\therefore (a + b)(a - b)=12$.$\because a - b = -2$,$\therefore a + b = -6$.
5. 分解因式:
(1)$x^2 - \frac{1}{4} =$
(2)$x(x - 2) + 1 =$
(3)$(x - 1)^2 - 9 =$
(1)$x^2 - \frac{1}{4} =$
$(x - \frac{1}{2})(x + \frac{1}{2})$
;(2)$x(x - 2) + 1 =$
$(x - 1)^2$
;(3)$(x - 1)^2 - 9 =$
$(x + 2)(x - 4)$
.答案
(1)$(x - \frac{1}{2})(x + \frac{1}{2})$;(2)$(x - 1)^2$;(3)$(x + 2)(x - 4)$
(2)$x(x - 2) + 1 = x^2 - 2x + 1 = (x - 1)^2$. (3)$(x - 1)^2 - 9 = (x - 1 + 3)(x - 1 - 3) = (x + 2)(x - 4)$.
(2)$x(x - 2) + 1 = x^2 - 2x + 1 = (x - 1)^2$. (3)$(x - 1)^2 - 9 = (x - 1 + 3)(x - 1 - 3) = (x + 2)(x - 4)$.
6. 用简便方法计算.
(1)$20.6^2 - 10.6^2$；(2)$204^2 + 204 × 192 + 96^2$.
(1)$20.6^2 - 10.6^2$；(2)$204^2 + 204 × 192 + 96^2$.
答案
解:(1)$20.6^2 - 10.6^2 = (20.6 + 10.6) × (20.6 - 10.6) = 31.2 × 10 = 312$.
(2)$204^2 + 204 × 192 + 96^2 = 204^2 + 204 × 2 × 96 + 96^2 = (204 + 96)^2 = 300^2 = 90\ 000$.
(2)$204^2 + 204 × 192 + 96^2 = 204^2 + 204 × 2 × 96 + 96^2 = (204 + 96)^2 = 300^2 = 90\ 000$.
7. 分解因式：$a^3 -9a = ( )$
A.$a(a-3)(a+3)$
B.$a(a^2 +9)$
C.$(a-3)(a+3)$
D.$a^2(a-9)$
A.$a(a-3)(a+3)$
B.$a(a^2 +9)$
C.$(a-3)(a+3)$
D.$a^2(a-9)$
答案
A $a^3 -9a = a(a^2 -9) = a(a + 3)(a - 3)$.
8. [2026长春153中期中]若$a + b = 3$，$ab = 1$，则$a^3b + 2a^2b^2 + ab^3$的值为
9
.答案
9 $\because a + b = 3$,$ab = 1$,$\therefore a^3b + 2a^2b^2 + ab^3 = ab(a^2 + 2ab + b^2) = ab(a + b)^2 = 1 × 3^2 = 9$.
9. 如图,若$R=6.75$,$r=3.25$,则图中阴影部分的面积为
35π
.(结果保留π)答案
35π 根据题意,得题图中阴影部分的面积为 $π R^2 - π r^2 = π(R^2 - r^2) = π(R + r)(R - r) = π × (6.75 + 3.25) × (6.75 - 3.25) = π × 10 × 3.5 = 35π$.
10. 教材P51例2变式 把下列多项式分解因式：
(1)$3x^2 - 18xy + 27y^2$；
(2)$x^2(x - y) + y^2(y - x)$。
(1)$3x^2 - 18xy + 27y^2$；
(2)$x^2(x - y) + y^2(y - x)$。
答案
解:(1)$3x^2 - 18xy + 27y^2 = 3(x^2 - 6xy + 9y^2) = 3(x - 3y)^2$.
(2)$x^2(x - y) + y^2(y - x) = x^2(x - y) - y^2(x - y) = (x - y)(x^2 - y^2) = (x - y)(x + y)(x - y) = (x - y)^2(x + y)$.
(2)$x^2(x - y) + y^2(y - x) = x^2(x - y) - y^2(x - y) = (x - y)(x^2 - y^2) = (x - y)(x + y)(x - y) = (x - y)^2(x + y)$.
11. 新趋势·代数推理 对于任意整数$m$,试说明: $(4m +5)^2 -9$ 总能被8整除.
答案
解:$(4m +5)^2 -9 = (4m +5)^2 - 3^2 = (4m +5 +3)(4m +5 -3) = (4m +8)(4m +2) = 4 × 2(m +2)(2m +1) = 8(m +2)(2m +1)$.
$\because m$ 是整数,$\therefore (m +2)(2m +1)$为整数,
$\therefore 8(m +2)(2m +1)$能被 8 整除,
$\therefore (4m +5)^2 -9$总能被 8 整除.
$\because m$ 是整数,$\therefore (m +2)(2m +1)$为整数,
$\therefore 8(m +2)(2m +1)$能被 8 整除,
$\therefore (4m +5)^2 -9$总能被 8 整除.
