《2025年初中必刷题八年级数学上册华师大版》

1 [2025 湖北黄石质检]计算$(-3a^{3}b)^{2}$的结果是( )

A.$9a^{5}b^{2}$
B.$9a^{6}b^{2}$
C.$6a^{9}b^{2}$
D.$-9a^{6}b^{2}$

答案： B　[解析]$(-3a^{3}b)^{2}=(-3)^{2}\cdot (a^{3})^{2}\cdot b^{2}=9a^{6}b^{2}$，故选B.

2 计算$(a^{2}+a^{2}+a^{2})^{3}$的结果是( )

A.$9a^{6}$
B.$27a^{5}$
C.$a^{18}$
D.$27a^{6}$

答案： D　[解析]$(a^{2}+a^{2}+a^{2})^{3}=(3a^{2})^{3}=27a^{6}$.故选D.

3 下列计算正确的是( )

A.$(xy)^{3}= xy^{3}$
B.$(-4xy^{2})^{2}= 16x^{2}y^{4}$
C.$(2xy)^{3}= 6x^{3}y^{3}$
D.$-(3x^{2})^{2}= 9x^{4}$

答案： B　[解析]A选项，$(xy)^{3}=x^{3}y^{3}$，故本选项不合题意；B选项，$(-4xy^{2})^{2}=16x^{2}y^{4}$，故本选项符合题意；C选项，$(2xy)^{3}=8x^{3}y^{3}$，故本选项不合题意；D选项，$-(3x^{2})^{2}=-9x^{4}$，故本选项不合题意.故选B.

4 [2024 河北石家庄期末]若$n,k$为正整数,则$(\underbrace{2 + 2 + … + 2}_{n个2})^{k}= $( )

A.$2^{k}n^{k}$
B.$k^{2n}$
C.$2n^{k}$
D.$2^{nk}$

答案： A　[解析]$\because \underbrace{2+2+\cdots +2}_{n个2}=2n$，$\therefore$ 原式$=(2n)^{k}=2^{k}n^{k}$，A选项符合题意，故选A.

5 已知$2^{n}= a,3^{n}= b,12^{n}= c$,那么$a,b,c$之间满足的等量关系是( )

A.$c = ab$
B.$c = ab^{2}$
C.$c = a^{2}b$
D.$c = a^{3}b$

答案： C　[解析]$\because 2^{n}=a$，$3^{n}=b$,$12^{n}=c$，$\therefore (4× 3)^{n}=c$，$\therefore 4^{n}× 3^{n}=c$，$(2^{2})^{n}× 3^{n}=c$，则$a^{2}b=c$，故选C.

6 已学的“幂的运算”有①同底数幂的乘法,②幂的乘方,③积的乘方.在“$(a^{2}\cdot a^{3})^{2}= (a^{2})^{2}(a^{3})^{2}= a^{4}\cdot a^{6}= a^{10}$”的运算过程中,运用了上述幂的运算中的______(按运算顺序填序号).

答案：
③②①　[解析]
　　　　

7 [2025 陕西宝鸡期中]若一个正方体的棱长为$4×10^{2}m$,则这个正方体的体积为______$m^{3}$.

答案： $6.4× 10^{7}$　[解析]$\because$ 一个正方体的棱长为$4× 10^{2}\ \text{m}$，$\therefore$ 它的体积是$(4× 10^{2})^{3}=6.4× 10^{7}\ (\text{m}^{3})$.故答案是$6.4× 10^{7}$.

8 [2024 江苏苏州期末]若$a^{m}= 4,b^{m}= 9$,则$(ab)^{m}$的值为______.

答案： 36　[解析]当$a^{m}=4$，$b^{m}=9$时，$(ab)^{m}=a^{m}\cdot b^{m}=4× 9=36$.故答案为36.

9 计算：(1)$(-3x^{3})^{2}-x^{2}\cdot x^{4}-(x^{2})^{3}$.
(2)$(-3a^{2})^{3}+(a^{3})^{2}+a^{2}\cdot a^{4}$.
(3)$(-2x^{2})^{3}+(-3x^{3})^{2}+x^{2}\cdot x^{4}$.
(4)$(-2a)^{6}-(-3a^{3})^{2}+[-(2a)^{2}]^{3}$.

答案： [解]
(1)原式$=9x^{6}-x^{6}-x^{6}=7x^{6}$.
(2)原式$=-27a^{6}+a^{6}+a^{6}=-25a^{6}$.
(3)原式$=-8x^{6}+9x^{6}+x^{6}=2x^{6}$.
(4)原式$=(-2)^{6}\cdot a^{6}-(-3)^{2}\cdot (a^{3})^{2}+(-1)^{3}\cdot (2a)^{6}=64a^{6}-9a^{6}-64a^{6}=-9a^{6}$.

10 [2024 湖南怀化期末]计算$(-\frac{1}{3})^{2023}×3^{2022}$的值是( )

A.$\frac{1}{3}$
B.$-\frac{1}{3}$
C.$\frac{1}{9}$
D.$-\frac{1}{9}$

答案： B　[解析]$\left(-\dfrac{1}{3}\right)^{2023}× 3^{2022}=\left(-\dfrac{1}{3}\right)^{2022}× \left(-\dfrac{1}{3}\right)× 3^{2022}=\left(-\dfrac{1}{3}× 3\right)^{2022}× \left(-\dfrac{1}{3}\right)=-\dfrac{1}{3}$，故选B.

11 已知$3^{3x + 1}×5^{3x + 1}= 15^{2x + 4}$,则$x = $______.

答案： 3　[解析]$\because 3^{3x+1}× 5^{3x+1}=15^{2x+4}$，$\therefore 15^{3x+1}=15^{2x+4}$，$\therefore 3x+1=2x+4$，$\therefore x=3$.

12 计算：
(1)$(-\frac{1}{3})^{2}-(\frac{2}{3})^{2021}×(-\frac{3}{2})^{2022}$.
(2)$(\frac{12}{5})^{11}×(-\frac{5}{6})^{13}×(\frac{1}{2})^{12}$.
(3)$(\frac{1}{10}×\frac{1}{9}×\frac{1}{8}×…×\frac{1}{2}×1)^{10}\cdot(10×9×8×…×2×1)^{10}$.

答案： [解]
(1)原式$=\dfrac{1}{9}+\left[\dfrac{2}{3}× \left(-\dfrac{3}{2}\right)\right]^{2021}× \dfrac{3}{2}=\dfrac{1}{9}-1× \dfrac{3}{2}=-\dfrac{25}{18}$.
(2)原式$=\left(-\dfrac{12}{5}× \dfrac{5}{6}× \dfrac{1}{2}\right)^{11}× \dfrac{1}{2}× \left(-\dfrac{5}{6}\right)^{2}=-\dfrac{1}{2}× \dfrac{25}{36}=-\dfrac{25}{72}$.
(3)$\left(\dfrac{1}{10}× \dfrac{1}{9}× \dfrac{1}{8}× \cdots × \dfrac{1}{2}× 1\right)^{10}\cdot (10× 9× 8× \cdots × 2× 1)^{10}=\left(\dfrac{1}{10}× \dfrac{1}{9}× \dfrac{1}{8}× \cdots × \dfrac{1}{2}× 1× 10× 9×8× \cdots × 2× 1\right)^{10}=1^{10}=1$.