《2025年轻松作业本八年级数学上册人教版》

8. (2024秋·烟台期中)在下列各式中:①$(\frac {-2n}{a^{2}b})^{2}$;②$-\frac {8m^{4}n^{2}}{a^{2}b}$;③$\frac {8m^{4}n^{2}}{a^{5}b}\cdot \frac {an}{bm^{2}}$;④$\frac {4n^{2}}{ab^{2}}÷a^{3}$,相等的两个式子是 (B)
A. ①②
B. ①④
C. ②③
D. ③④

答案： B

9. (1)(2024秋·娄底期中)计算:$(-\frac {b}{2a})^{2}\cdot (\frac {3a}{b})^{3}÷\frac {a^{2}}{4b}=$$\frac{27}{a}$；
(2) 计算$\frac {x^{2}-2xy+y^{2}}{x^{2}}÷\frac {x-y}{x}$的结果为$\frac{x - y}{x}$.

答案：
(1) $\frac{27}{a}$
(2) $\frac{x - y}{x}$

10. 计算:
(1)$(-\frac {x}{y})^{2}\cdot (-\frac {x^{2}}{y^{3}})^{3}÷(-\frac {x}{y})^{4}=$$-\frac{x^{4}}{y^{7}}$；
(2)$(-\frac {a}{b})^{2}\cdot (-\frac {b}{a})^{3}÷(-ab^{4})=$$\frac{1}{a^{2}b^{3}}$.

答案：
(1) $-\frac{x^{4}}{y^{7}}$
(2) $\frac{1}{a^{2}b^{3}}$

11. 先化简,再求值:$[\frac {xy}{(x-y)^{2}}]^{2}\cdot (\frac {x-y}{xy^{2}})^{2}÷(\frac {1}{xy-y^{2}})^{3}$,其中$x= -2,y= 4$.
解：$[\frac{xy}{(x - y)^{2}}]^{2} \cdot (\frac{x - y}{xy^{2}})^{2} ÷ (\frac{1}{xy - y^{2}})^{3} = \frac{(xy)^{2}}{(x - y)^{4}} \cdot \frac{(x - y)^{2}}{x^{2}y^{4}} ÷ \frac{1}{[y(x - y)]^{3}} = \frac{x^{2}y^{2}}{(x - y)^{4}} \cdot \frac{(x - y)^{2}}{x^{2}y^{4}} ÷ \frac{1}{y^{3}(x - y)^{3}} = \frac{1}{y^{2}(x - y)^{2}} × y^{3}(x - y)^{3} = $$xy - y^{2}$，当 $x = -2$，$y = 4$ 时，原式 $= (-2) × 4 - 4^{2} = $$-24$。

答案： 解：$[\frac{xy}{(x - y)^{2}}]^{2} \cdot (\frac{x - y}{xy^{2}})^{2} \div (\frac{1}{xy - y^{2}})^{3} = \frac{(xy)^{2}}{(x - y)^{4}} \cdot \frac{(x - y)^{2}}{x^{2}y^{4}} \div \frac{1}{[y(x - y)]^{3}} = \frac{x^{2}y^{2}}{(x - y)^{4}} \cdot \frac{(x - y)^{2}}{x^{2}y^{4}} \div \frac{1}{y^{3}(x - y)^{3}} = \frac{1}{y^{2}(x - y)^{2}} \times y^{3}(x - y)^{3} = xy - y^{2}$，当 $x = -2$，$y = 4$ 时，原式 $= (-2) \times 4 - 4^{2} = -8 - 16 = -24$。

12. 已知$A= \frac {x^{2}}{x^{2}-2xy}÷\frac {x^{2}-4y^{2}}{x^{2}-4xy+4y^{2}}$.
(1) 化简A；
(2) 若$x^{2}-6xy+9y^{2}= 0$,求A的值.
(1) $\frac{x}{x + 2y}$；
(2) $\frac{3}{5}$.

答案： 解：
(1) $A = \frac{x^{2}}{x^{2} - 2xy} \div \frac{x^{2} - 4y^{2}}{x^{2} - 4xy + 4y^{2}} = \frac{x^{2}}{x(x - 2y)} \times \frac{(x - 2y)^{2}}{(x + 2y)(x - 2y)} = \frac{x}{x + 2y}$；
(2) $\because x^{2} - 6xy + 9y^{2} = 0$，$\therefore (x - 3y)^{2} = 0$。$\therefore x - 3y = 0$。$\therefore x = 3y$。$\therefore A = \frac{3y}{3y + 2y} = \frac{3y}{5y} = \frac{3}{5}$。

13. 甲、乙两人分别从相距s km的两地同时出发,若同向而行,经过$m_{1}$h甲追上乙;若相向而行,经过$m_{2}$h甲、乙两人相遇.设甲的平均速度为$v_{1}km/h$,乙的平均速度为$v_{2}km/h$,那么$\frac {v_{1}}{v_{2}}$等于多少？(用含$m_{1}$、$m_{2}$的式子表示,并说明理由)

答案： 解：依题意，得：$m_{1}v_{1} - m_{1}v_{2} = s$，$m_{2}v_{1} + m_{2}v_{2} = s$，$\therefore m_{1}v_{1} - m_{1}v_{2} = m_{2}v_{1} + m_{2}v_{2}$。$\therefore m_{1}v_{1} - m_{2}v_{1} = m_{1}v_{2} + m_{2}v_{2}$，$(m_{1} - m_{2})v_{1} = (m_{1} + m_{2})v_{2}$。$\therefore \frac{v_{1}}{v_{2}} = \frac{m_{1} + m_{2}}{m_{1} - m_{2}}$。