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1. [2025德州中考]已知$m, n$是正整数,且满足$3^m · 3^m · 3^m = 3^n$,则$m$与$n$的关系正确的是 (
A.$3m = n$
B.$m^3 = n$
C.$m + 3 = n$
D.$m + 1 = n$
A
)A.$3m = n$
B.$m^3 = n$
C.$m + 3 = n$
D.$m + 1 = n$
答案
1 A $\because 3^m · 3^m · 3^m = 3^n,\therefore 3^{3m} = 3^n,\therefore 3m = n.$
2. [2025无锡中考]分解因式$a^3 -4a$的结果是（
A.$a(a^2 +4)$
B.$a(a -4)$
C.$a(a +2)(a -2)$
D.$a(a^2 -1)$
C
）A.$a(a^2 +4)$
B.$a(a -4)$
C.$a(a +2)(a -2)$
D.$a(a^2 -1)$
答案
2 C $a^3 -4a = a(a^2 -4) = a(a +2)(a -2).$
3. [2025内江中考] 下列计算正确的是（
A.$x^2 · x^4 = x^8$
B.$(x - y)^2 = x^2 - y^2$
C.$x + 2x^2 = 3x^2$
D.$(x + 2)(x - 2) = x^2 - 4$
D
）A.$x^2 · x^4 = x^8$
B.$(x - y)^2 = x^2 - y^2$
C.$x + 2x^2 = 3x^2$
D.$(x + 2)(x - 2) = x^2 - 4$
答案
3 D 选项A,$x^2 · x^4 = x^6$;选项B,$(x - y)^2 = x^2 - 2xy + y^2$;选项C,$x$与$2x^2$不是同类项,无法合并.
4. [2025自贡中考]若$2a + b = -1$，则$4a^2 + 2ab - b$的值为
1
。答案
4 1 $\because 2a + b = -1,\therefore b = -1 -2a,\therefore 4a^2 + 2ab - b =4a^2 + b(2a -1) =4a^2 + (-1 -2a)(2a -1) =4a^2 -(2a +1)(2a -1) =4a^2 -(4a^2 -1) =1.$
5. [2025哈尔滨中考]定义新运算：$a\otimes b=2ab - b^2$，则$(3n)\otimes(2n)$的运算结果是
$8n^2$
。答案
5 $8n^2$ $\because a\otimes b = 2ab - b^2,\therefore (3n)\otimes(2n) = 2 × 3n · 2n -(2n)^2 =12n^2 -4n^2 =8n^2.$
6. [2025潍坊中考] 先化简,再求值:$x(5x - 8y) - 4(x - y)^2$,其中$x,y$满足$x + 2y = 0.$
答案
6 解:$x(5x - 8y) - 4(x - y)^2 =5x^2 -8xy -4(x^2 -2xy + y^2) =5x^2 -8xy -4x^2 +8xy -4y^2 =x^2 -4y^2.$
$\because x + 2y =0,\therefore x^2 -4y^2 =x^2 -(2y)^2 =(x + 2y)(x - 2y) =0.$
$\because x + 2y =0,\therefore x^2 -4y^2 =x^2 -(2y)^2 =(x + 2y)(x - 2y) =0.$
7. [2023河北中考]现有甲、乙、丙三种长方形和正方形卡片各若干张,卡片的边长如图1所示$(a＞1)$.某同学分别用6张卡片拼出了两个长方形(不重叠无缝隙),如图2和图3所示,其面积分别为$S_1,S_2$.
(1)请用含$a$的式子分别表示$S_1,S_2$,当$a=2$时,求$S_1+S_2$的值;
(2)比较$S_1$与$S_2$的大小,并说明理由.
(1)请用含$a$的式子分别表示$S_1,S_2$,当$a=2$时,求$S_1+S_2$的值;
(2)比较$S_1$与$S_2$的大小,并说明理由.
答案
7 解:(1)根据题意,得三种卡片的面积分别为$S_甲 =a^2,S_乙 =a,S_丙 =1$,
$\therefore S_1 = S_甲 +3S_乙 +2S_丙 =a^2 +3a +2,S_2 =5S_乙 +S_丙 =5a +1$,
$\therefore S_1 + S_2 =(a^2 +3a +2) +(5a +1) =a^2 +8a +3.$
当$a=2$时,$S_1 + S_2 =2^2 +8 ×2 +3 =23.$
(2)$S_1 ＞ S_2$.理由如下:
$\because S_1 =a^2 +3a +2,S_2 =5a +1$,
$\therefore S_1 - S_2 =(a^2 +3a +2) -(5a +1) =a^2 -2a +1 =(a -1)^2.$
$\because a ＞1,\therefore (a -1)^2 ＞0,\therefore S_1 ＞ S_2.$
$\therefore S_1 = S_甲 +3S_乙 +2S_丙 =a^2 +3a +2,S_2 =5S_乙 +S_丙 =5a +1$,
$\therefore S_1 + S_2 =(a^2 +3a +2) +(5a +1) =a^2 +8a +3.$
当$a=2$时,$S_1 + S_2 =2^2 +8 ×2 +3 =23.$
(2)$S_1 ＞ S_2$.理由如下:
$\because S_1 =a^2 +3a +2,S_2 =5a +1$,
$\therefore S_1 - S_2 =(a^2 +3a +2) -(5a +1) =a^2 -2a +1 =(a -1)^2.$
$\because a ＞1,\therefore (a -1)^2 ＞0,\therefore S_1 ＞ S_2.$
8. [2024福建中考]已知实数 $a,b,c,m,n$ 满足 $3m + n = \dfrac{b}{a}, mn = \dfrac{c}{a}$.
(1) 试说明: $b^2 -12ac$ 为非负数.
(2) 若 $a,b,c$ 均为奇数，$m,n$ 是否可以都为整数？说明你的理由.
(1) 试说明: $b^2 -12ac$ 为非负数.
(2) 若 $a,b,c$ 均为奇数，$m,n$ 是否可以都为整数？说明你的理由.
答案
8 解题思路:(2)分情况讨论:①$m,n$ 都为奇数;②$m,n$ 都为整数,且其中至少有一个为偶数.
解:(1)$\because 3m +n = \dfrac{b}{a},mn = \dfrac{c}{a}$,
$\therefore b =a(3m +n),c =amn$,
$\therefore b^2 -12ac =[a(3m +n)]^2 -12a^2 mn$
$=a^2(9m^2 +6mn +n^2) -12a^2 mn$
$=a^2(9m^2 -6mn +n^2)$
$=a^2(3m -n)^2.$
$\because a,m,n$ 是实数,$\therefore a^2(3m -n)^2 ≥ 0$,
$\therefore b^2 -12ac$ 为非负数.
(2)若$a,b,c$ 均为奇数,则$m,n$ 不可以都为整数.理由如下:
①当$m,n$ 均为奇数时,则$3m +n$ 为偶数,
$\therefore b =a(3m +n)$ 是一个偶数,
这与$b$ 为奇数相矛盾.
②当$m,n$ 均为偶数时,则$3m +n$ 为偶数,
$\therefore b =a(3m +n)$ 是一个偶数,
这与$b$ 为奇数相矛盾.
③当$m,n$ 中有一个为奇数,另一个为偶数时,可知$mn$ 为偶数.
由 $mn = \dfrac{c}{a}$,得 $c =amn$.
$\because amn$ 是一个偶数,这与$c$ 是奇数相矛盾.
综上所述,$m,n$ 不可以都为整数.
解:(1)$\because 3m +n = \dfrac{b}{a},mn = \dfrac{c}{a}$,
$\therefore b =a(3m +n),c =amn$,
$\therefore b^2 -12ac =[a(3m +n)]^2 -12a^2 mn$
$=a^2(9m^2 +6mn +n^2) -12a^2 mn$
$=a^2(9m^2 -6mn +n^2)$
$=a^2(3m -n)^2.$
$\because a,m,n$ 是实数,$\therefore a^2(3m -n)^2 ≥ 0$,
$\therefore b^2 -12ac$ 为非负数.
(2)若$a,b,c$ 均为奇数,则$m,n$ 不可以都为整数.理由如下:
①当$m,n$ 均为奇数时,则$3m +n$ 为偶数,
$\therefore b =a(3m +n)$ 是一个偶数,
这与$b$ 为奇数相矛盾.
②当$m,n$ 均为偶数时,则$3m +n$ 为偶数,
$\therefore b =a(3m +n)$ 是一个偶数,
这与$b$ 为奇数相矛盾.
③当$m,n$ 中有一个为奇数,另一个为偶数时,可知$mn$ 为偶数.
由 $mn = \dfrac{c}{a}$,得 $c =amn$.
$\because amn$ 是一个偶数,这与$c$ 是奇数相矛盾.
综上所述,$m,n$ 不可以都为整数.
