2025年新编基础训练八年级数学上册人教版第160页答案
4. 计算$\frac{x^2}{y} ÷ \frac{y}{x} \cdot (\frac{y}{x})^2$的结果是(
A
)
A. $x$ B. $x^2$ C. $y^2$ D. $y$
5. 化简:
(1) $(-ab^2)^2 ÷ (-\frac{3a^3b^2}{c}) = $
$-\frac{b^2c}{3a}$

(2) $(-\frac{x}{y^2})^2 \cdot (-\frac{y^2}{x})^3 ÷ (-\frac{y}{x})^4 = $
$-\frac{x^3}{y^2}$

6. 计算:
(1) $(\frac{-a}{b})^2 ÷ (\frac{2a^2}{5b})^2 \cdot (-\frac{a}{5b})$;
$-\frac{5}{4ab}$

(2) $(\frac{a^3b^2}{2c})^2 \cdot (-\frac{bc}{a^2})^3 ÷ (-\frac{3b^2c}{2a})^2$;
$-\frac{a^2b^3}{9c}$

(3) $\frac{x^2 - 4y^2}{x^2 + 2xy + y^2} ÷ (\frac{x + 2y}{x + y})^2$。
$\frac{x-2y}{x+2y}$

答案

4. A
5. (1) $-\frac{b^2c}{3a}$;(2) $-\frac{x^3}{y^2}$
6. (1) $-\frac{5}{4ab}$;(2) $-\frac{a^2b^3}{9c}$;(3) $\frac{x-2y}{x+2y}$

解析

4. 原式=$\frac{x^2}{y} × \frac{x}{y} × \frac{y^2}{x^2}=\frac{x^3 y^2}{x^2 y^2}=x$
5. (1) 原式$=a^2b^4 × (-\frac{c}{3a^3b^2})=-\frac{b^2c}{3a}$;(2) 原式$=\frac{x^2}{y^4} × (-\frac{y^6}{x^3}) × \frac{x^4}{y^4}=-\frac{x^3}{y^2}$
6. (1) 原式$=\frac{a^2}{b^2} × \frac{25b^2}{4a^4} × (-\frac{a}{5b})=-\frac{5}{4ab}$;(2) 原式$=\frac{a^6b^4}{4c^2} × (-\frac{b^3c^3}{a^6}) × \frac{4a^2}{9b^4c^2}=-\frac{a^2b^3}{9c}$;(3) 原式$=\frac{(x-2y)(x+2y)}{(x+y)^2} × \frac{(x+y)^2}{(x+2y)^2}=\frac{x-2y}{x+2y}$
1. 计算$\frac{y}{x} ÷ \frac{y}{2} \cdot \frac{2}{y}$的结果是(
A
)
A. $\frac{4}{xy}$ B. $\frac{1}{2}x$ C. $\frac{y}{x}$ D. $2y$
2. 计算$(\frac{3y}{-2x})^2$的结果是(
C
)
A. $\frac{3y^2}{2x^2}$ B. $-\frac{3y^2}{2x^2}$ C. $\frac{9y^2}{4x^2}$ D. $-\frac{9y^2}{4x^2}$
3. 下列计算结果正确的是(
B
)
A. $(\frac{2a}{b})^2 = \frac{2a^2}{b^2}$ B. $(-\frac{2a}{b^2})^3 = -\frac{8a^3}{b^6}$
C. $(-\frac{a}{b^2})^3 = \frac{a^3}{b^6}$ D. $(-\frac{a^3}{b^2})^2 = -\frac{a^6}{b^4}$
4. 计算$(-\frac{2a}{b})^3 \cdot (\frac{2b}{a})^2 ÷ (-\frac{2b}{a})^2$的结果是(
B
)
A. $-\frac{8a}{b^6}$ B. $-\frac{8a^3}{b^3}$ C. $\frac{16a^2}{b^6}$ D. $-\frac{16a^2}{b^6}$
5. 计算:$(\frac{-2a^2b}{3c})^2 ÷ (\frac{2ab}{3c})^3 \cdot \frac{a}{c} = $
$\frac{3a^2}{2b}$

6. 计算:(1) $\frac{ab^2}{6c^2} \cdot \frac{-4c}{b^2} ÷ \frac{a}{c}$;
(2) $\frac{x^2 - 4}{x + 2} ÷ (x - 2) \cdot \frac{1}{x - 2}$。
(1)$-\frac{2c}{3}$;(2)$\frac{1}{x-2}$

7. 已知$\frac{1}{a} - 1 = 0$,求$(\frac{a - 3}{a^2 + 3a} ÷ \frac{a - 1}{a^2 + 6a + 9}) ÷ \frac{a + 3}{a - 1}$的值。
$-2$

8. 计算:
(1) $\frac{x^2}{y} ÷ (\frac{-y^2}{x}) \cdot (\frac{y}{-x})^2$;
(2) $6a^2b ÷ (-\frac{a}{2b})^2 \cdot \frac{a}{4b^2}$;
(3) $(\frac{2xz}{y})^2 ÷ (-\frac{xz}{3y}) \cdot \frac{8x}{y^3}$;
(4) $(\frac{a - 3}{a})^2 \cdot \frac{a^2 + a}{a - 3} \cdot \frac{a}{a + 1}$。
(1)$-\frac{x}{y}$;(2)$6ab$;(3)$-\frac{96x^2z}{y^4}$;(4)$a-3$

答案

【解析】:
1. $\frac{y}{x} ÷ \frac{y}{2} \cdot \frac{2}{y} = \frac{y}{x} \cdot \frac{2}{y} \cdot \frac{2}{y} = \frac{4}{xy}$
【答案】:A
【解析】:
2. $(\frac{3y}{-2x})^2 = \frac{(3y)^2}{(-2x)^2} = \frac{9y^2}{4x^2}$
【答案】:C
【解析】:
3. A.$(\frac{2a}{b})^2 = \frac{4a^2}{b^2}$,错误;B.$(-\frac{2a}{b^2})^3 = -\frac{8a^3}{b^6}$,正确;C.$(-\frac{a}{b^2})^3 = -\frac{a^3}{b^6}$,错误;D.$(-\frac{a^3}{b^2})^2 = \frac{a^6}{b^4}$,错误
【答案】:B
【解析】:
4. $(-\frac{2a}{b})^3 \cdot (\frac{2b}{a})^2 ÷ (-\frac{2b}{a})^2 = -\frac{8a^3}{b^3} \cdot \frac{4b^2}{a^2} \cdot \frac{a^2}{4b^2} = -\frac{8a^3}{b^3}$
【答案】:B
【解析】:
5. $(\frac{-2a^2b}{3c})^2 ÷ (\frac{2ab}{3c})^3 \cdot \frac{a}{c} = \frac{4a^4b^2}{9c^2} \cdot \frac{27c^3}{8a^3b^3} \cdot \frac{a}{c} = \frac{3a^2}{2b}$
【答案】:$\frac{3a^2}{2b}$
【解析】:
6.(1)$\frac{ab^2}{6c^2} \cdot \frac{-4c}{b^2} ÷ \frac{a}{c} = \frac{ab^2}{6c^2} \cdot \frac{-4c}{b^2} \cdot \frac{c}{a} = -\frac{2c}{3}$
(2)$\frac{x^2 - 4}{x + 2} ÷ (x - 2) \cdot \frac{1}{x - 2} = \frac{(x-2)(x+2)}{x+2} \cdot \frac{1}{x-2} \cdot \frac{1}{x-2} = \frac{1}{x-2}$
【答案】:(1)$-\frac{2c}{3}$;(2)$\frac{1}{x-2}$
【解析】:
7. 由$\frac{1}{a}-1=0$得$a=1$。原式$=(\frac{a-3}{a(a+3)} \cdot \frac{(a+3)^2}{a-1}) \cdot \frac{a-1}{a+3} = \frac{a-3}{a}$,代入$a=1$得$-2$
【答案】:$-2$
【解析】:
8.(1)$\frac{x^2}{y} ÷ (-\frac{y^2}{x}) \cdot (\frac{y}{-x})^2 = \frac{x^2}{y} \cdot (-\frac{x}{y^2}) \cdot \frac{y^2}{x^2} = -\frac{x}{y}$
(2)$6a^2b ÷ (-\frac{a}{2b})^2 \cdot \frac{a}{4b^2} = 6a^2b \cdot \frac{4b^2}{a^2} \cdot \frac{a}{4b^2} = 6ab$
(3)$(\frac{2xz}{y})^2 ÷ (-\frac{xz}{3y}) \cdot \frac{8x}{y^3} = \frac{4x^2z^2}{y^2} \cdot (-\frac{3y}{xz}) \cdot \frac{8x}{y^3} = -\frac{96x^2z}{y^4}$
(4)$(\frac{a-3}{a})^2 \cdot \frac{a^2+a}{a-3} \cdot \frac{a}{a+1} = \frac{(a-3)^2}{a^2} \cdot \frac{a(a+1)}{a-3} \cdot \frac{a}{a+1} = a-3$
【答案】:(1)$-\frac{x}{y}$;(2)$6ab$;(3)$-\frac{96x^2z}{y^4}$;(4)$a-3$

解析

1. A
2. C
3. B
4. B
5. $\frac{3a^{2}}{2c^{3}}$
6.
(1) $-\frac{2}{3}$
(2) $\frac{1}{x - 2}$
7. $-2$
8.
(1) $-\frac{x^{2}}{y^{3}}$
(2) $6a$
(3) $-\frac{144x^{2}}{y^{4}}$
(4) $a - 3$