《2025年一遍过八年级数学上册华师大版》

1 计算$x^{4}÷x$的结果正确的是(B)
A.$x^{4}$
B.$x^{3}$
C.$x^{2}$
D.$x$

答案： B $x^{4}÷ x=x^{4 - 1}$（指数相减）$=x^{3}$.

2 $(-a^{6})÷(-a)^{2}$的运算结果是(B)

A.$a^{4}$
B.$-a^{4}$
C.$a^{3}$
D.$-a^{3}$
[变式] 计算：$-a^{10}÷(-a)^{5}·(-a)^{5}= $$-a^{10}$.

答案： B $(-a^{6})÷ (-a)^{2}=-a^{6}÷ a^{2}=-(a^{6}÷ a^{2})=-a^{4}$.
@@$-a^{10}$ $-a^{10}÷ (-a)^{5}\cdot (-a)^{5}=-a^{10}÷ a^{5}\cdot a^{5}=-a^{10 - 5 + 5}=-a^{10}$.

3 计算$(a^{3})^{2}÷a^{2}$的结果是(B)
A.$a^{3}$
B.$a^{4}$
C.$a^{7}$
D.$a^{8}$

答案： B $(a^{3})^{2}÷ a^{2}=a^{3× 2}÷ a^{2}=a^{6 - 2}=a^{4}$.

4 [2024 山西中考]下列运算正确的是(D)
A.$2m + n = 2mn$
B.$m^{6}÷m^{2}= m^{3}$
C.$(-mn)^{2}= -m^{2}n^{2}$
D.$m^{2}·m^{3}= m^{5}$

答案： D A选项，2m和n不是同类项，不能合并；B选项，$m^{6}÷ m^{2}=m^{4}$；C选项，$(-mn)^{2}=m^{2}n^{2}$.

5 [2025 杭州期末]$2025^{5}是2025^{4}$的(B)
A.5 倍
B.2025 倍
C.$\dfrac{5}{4}$倍
D.1 倍

答案： B $2025^{5}÷ 2025^{4}=2025$倍.

6 若$(x^{2})^{3}÷x^{m}= x^{4}$,则$m=$2.
[变式] 若$a^{9}÷a^{x - 2}= (-a)^{6}$,则$x$的值为5.

答案： 2 $\because (x^{2})^{3}÷ x^{m}=x^{4}$，$\therefore x^{6}÷ x^{m}=x^{6 - m}=x^{4}$，$\therefore 6 - m = 4$，$\therefore m = 2$.
@@5 $\because a^{9}÷ a^{x - 2}=(-a)^{6}=a^{6}$，$\therefore 9 - (x - 2)=6$，$\therefore x = 5$.

7 已知$x - y = 2$,则$2^{x}÷2^{y}= $4.

答案： 4 $\because x - y = 2$，$\therefore 2^{x}÷ 2^{y}=2^{x - y}=2^{2}=4$.

8 计算：(1)$a^{5}= a^{10}÷$$a^{5}$$=$______$a^{8}$$÷a^{3}$；
(2)$x^{14}÷(x^{10}÷x^{2})= $$x^{6}$.

答案：
(1)$a^{5}$ $a^{8}$；
(2)$x^{6}$
(2)$x^{14}÷ (x^{10}÷ x^{2})=x^{14}÷ x^{8}=x^{6}$.

9 [2025 武汉期末]若$x^{a}= 3$,$x^{b}= 5$,则$x^{a + b}$的值是15,$x^{2a - b}$的值是$\frac{9}{5}$.

答案： 15 $\frac{9}{5}$ $x^{a + b}=x^{a}\cdot x^{b}=3× 5 = 15$，$x^{2a - b}=(x^{a})^{2}÷ x^{b}=3^{2}÷ 5=\frac{9}{5}$.

10 [一题多解][2025 上海期末]计算：$8^{6}×(-\dfrac{1}{2})^{17}$.

答案： 解：通解 $8^{6}× (-\frac{1}{2})^{17}=8^{6}× (-\frac{1}{2^{17}})=-8^{6}× \frac{1}{2^{17}}=-\frac{8^{6}}{2^{17}}=-\frac{262144}{131072}=-2$.优解1（同底数幂的除法）$8^{6}× (-\frac{1}{2})^{17}=(2^{3})^{6}× (-\frac{1}{2^{17}})=2^{18}× (-\frac{1}{2^{17}})=-\frac{2^{18}}{2^{17}}=-2$.优解2（积的乘方的运算法则的逆用）$8^{6}× (-\frac{1}{2})^{17}=8^{6}× (-\frac{1}{2})^{6}× (-\frac{1}{2})^{6}× (-\frac{1}{2})^{5}=8^{6}× (\frac{1}{2})^{6}× (\frac{1}{2})^{6}× (-\frac{1}{2})^{5}=(8× \frac{1}{2}× \frac{1}{2})^{6}× (-\frac{1}{2})^{5}=2^{6}× (-\frac{1}{2})^{5}=2× 2^{5}× (-\frac{1}{2})^{5}=2× (-\frac{1}{2}× 2)^{5}=2× (-1)=-2$.优解3（幂的乘方与积的乘方的运算法则的逆用）$8^{6}× (-\frac{1}{2})^{17}=(2^{3})^{6}× (-\frac{1}{2})^{17}=2^{18}× (-\frac{1}{2})^{17}=2^{17}× (-\frac{1}{2})^{17}× 2=(-2× \frac{1}{2})^{17}× 2=(-1)× 2=-2$.

11 [一题多解][2025 吉林五中期中]若$2a - 3b + c - 2 = 0$,求$16^{a}÷8^{2b}×4^{c}$的值.

答案： 解：通解1 $\because 2a - 3b + c - 2 = 0$，$\therefore 2a - 3b + c = 2$，$\therefore (2^{4})^{a}÷ (2^{3})^{2b}× (2^{2})^{c}=2^{4a}÷ 2^{6b}× 2^{2c}=2^{4a - 6b + 2c}=2^{2(2a - 3b + c)}=2^{2× 2}=2^{4}=16$.通解2 $\because 2a - 3b + c - 2 = 0$，$\therefore 2a - 3b + c = 2$，$\therefore 16^{a}÷ 8^{2b}× 4^{c}=4^{2a}÷ 64^{b}× 4^{c}=4^{2a}÷ 4^{3b}× 4^{c}=4^{2a - 3b + c}=4^{2}=16$.

12 [2025 常德期末]若$5^{m}= 3$,$25^{n}= 4$,求$5^{3m - 4n}$的值.
[变式] 已知$4^{m}= 5$,$8^{n}= 3$,$3^{m}= 4$.
(1)求$2^{2m + 3n}$的值；
(2)求$2^{4m - 6n}$的值；
(3)[一题多解]求$12^{2m}$的值.

答案： 解：$\because 5^{m}=3$，$25^{n}=4$，$\therefore (5^{2})^{n}=5^{2n}=4$，$\therefore 5^{3m - 4n}=5^{3m}÷ 5^{4n}=(5^{m})^{3}÷ (5^{2n})^{2}=3^{3}÷ 4^{2}=\frac{27}{16}$.
@@解：
(1)$\because 4^{m}=2^{2m}=5$，$8^{n}=2^{3n}=3$，$3^{m}=4$，$\therefore 2^{2m + 3n}=2^{2m}\cdot 2^{3n}=5× 3 = 15$.
(2)$2^{4m - 6n}=2^{4m}÷ 2^{6n}=(2^{2m})^{2}÷ (2^{3n})^{2}=5^{2}÷ 3^{2}=\frac{25}{9}$.
(3)通解1 $12^{2m}=(3× 4)^{2m}=3^{2m}× 4^{2m}=(3^{m})^{2}× (4^{m})^{2}=4^{2}× 5^{2}=16× 25 = 400$.通解2 $12^{2m}=[(3× 4)^{m}]^{2}=(3^{m}× 4^{m})^{2}=(4× 5)^{2}=20^{2}=400$.

13 已知$25^{a}·5^{2b}= 5^{6}$,$4^{b}÷4^{c}= 4$,求代数式$a + 2b - c$的值.

答案： 解：$\because 25^{a}\cdot 5^{2b}=5^{6}$，$4^{b}÷ 4^{c}=4$，$\therefore 5^{2a}\cdot 5^{2b}=5^{6}$，$4^{b - c}=4$，$\therefore 2a + 2b = 6$，$b - c = 1$，$\therefore a + b = 3$，$\therefore a + 2b - c = a + b + b - c = 3 + 1 = 4$.