《2025年名校课堂八年级数学上册人教版安徽专版》

1. (2023·六安金安区期末)化简 $ x ÷ \frac{x}{y} \cdot \frac{1}{x} $ 的结果为 (B)

A.$ \frac{x}{y} $
B.$ \frac{y}{x} $
C.$ xy $
D.1

答案： B

2. 计算: $ \frac{n}{m^{2}} ÷ \frac{n^{2}}{m^{3}} ÷ \frac{m}{n^{2}} = $n.

答案： n

3. 计算:
(1) $ \frac{3b^{2}}{16a} \cdot \frac{2a}{b} ÷ \frac{bc}{2a^{2}} $.
(2) $ (-2ab) ÷ \frac{ab^{2}}{a - b} \cdot \frac{1}{2(a - b)^{2}} $.

答案： 解:
(1)原式$=\frac {3b^{2}}{16a}\cdot \frac {2a}{b}\cdot \frac {2a^{2}}{bc}=\frac {3a^{2}}{4c}$.
(2)原式$=(-2ab)\cdot \frac {a-b}{ab^{2}}\cdot \frac {1}{2(a-b)^{2}}$$=-\frac {1}{b(a-b)}=\frac {1}{b^{2}-ab}.$

4. 计算:
(1) $ (\frac{2x^{2}}{3y})^{2} = \frac{(2x^{2})^{2}}{(3y)^{2}} = $____.
(2) $ (-\frac{y^{2}}{2x^{3}})^{3} = -\frac{(y^{2})^{3}}{(2x^{3})^{3}} = $-\frac{y^{6}}{8x^{9}}.

答案：
(1)$2x^{2}$ $3y$ $\frac{4x^{4}}{9y^{2}}$
(2)$y^{2}$ $2x^{3}$ $-\frac{y^{6}}{8x^{9}}$

5. 计算:
(1) $ (\frac{-y^{2}}{x})^{2} $. (2) $ (\frac{2a^{2}b}{-c^{2}})^{3} $.

答案： 解:
(1)原式$=\frac {(-y^{2})^{2}}{x^{2}}=\frac {y^{4}}{x^{2}}$.
(2)原式$=\frac {(2a^{2}b)^{3}}{(-c^{2})^{3}}=-\frac {8a^{6}b^{3}}{c^{6}}.$

6. (2023·河北)化简 $ x^{3} \cdot (\frac{y^{3}}{x})^{2} $ 的结果是 (A)

A.$ xy^{6} $
B.$ xy^{5} $
C.$ x^{2}y^{5} $
D.$ x^{2}y^{6} $

答案： A

7. 计算:
(1) $ (\frac{2xy^{3}}{z^{2}})^{2} ÷ \frac{6x^{2}}{y^{3}} $.
(2) $ -(\frac{m}{n})^{2} \cdot (-\frac{n^{2}}{m})^{4} ÷ (-mn^{4}) $.
(3) $ (-\frac{b^{2}}{2a})^{2} ÷ (\frac{b}{a^{2}})^{3} \cdot \frac{b}{2} $.

答案： 解:
(1)原式$=\frac {4x^{2}y^{6}}{z^{4}}\cdot \frac {y^{3}}{6x^{2}}=\frac {2y^{9}}{3z^{4}}$.
(2)原式$=\frac {m^{2}}{n^{2}}\cdot \frac {n^{8}}{m^{4}}\cdot \frac {1}{mn^{4}}=\frac {n^{2}}{m^{3}}$.
(3)原式$=\frac {b^{4}}{4a^{2}}÷\frac {b^{3}}{a^{6}}\cdot \frac {b}{2}=\frac {b^{4}}{4a^{2}}\cdot \frac {a^{6}}{b^{3}}\cdot \frac {b}{2}=\frac {a^{4}b^{2}}{8}.$

8. 先化简, 再求值: $ (\frac{1}{x - 1})^{2} \cdot \frac{2x + 6}{x^{2} - 1} ÷ \frac{x + 3}{x^{2} - 2x + 1} $, 其中 $ x = 2 $.

答案： 解:原式$=\frac {1}{(x-1)^{2}}\cdot \frac {2(x+3)}{(x+1)(x-1)}\cdot \frac {(x-1)^{2}}{x+3}=\frac {2}{(x+1)(x-1)}$.当$x=2$时,原式$=\frac {2}{3}.$