零五网 全部参考答案 启东中学作业本 2025年启东中学作业本八年级数学上册人教版 第25页解析答案
计算:
(1)$(\frac {-2a^{2}b}{3c})^{2};$ (2)$(\frac {-3x^{2}y}{2x})^{3};$
(3)$\frac {4x}{3y}\cdot \frac {y}{2x^{4}}÷(\frac {1}{x})^{2};$ (4)$(-\frac {a}{b})^{2}÷\frac {3a}{4b}\cdot \frac {2b}{3a};$
(5)$(\frac {-3ac}{2b})^{2}÷(-9ac^{2});$ (6)$(\frac {a^{2}b}{-c})^{3}\cdot (\frac {c^{2}}{-ab})^{2}÷(\frac {bc}{a})^{4};$
(7)$(-\frac {a}{b})^{2}\cdot (-\frac {a}{b})^{3}÷(-a^{4}b);$ (8)$(\frac {2ab^{3}}{-c^{2}d})^{2}÷\frac {6a^{2}}{b^{3}}\cdot (\frac {-3c}{b^{2}})^{3};$
(9)$6xy^{2}÷(-\frac {3y^{2}}{2x})^{3}\cdot (\frac {2y}{3x})^{2};$ (10)$(\frac {-y^{3}}{x})^{2}\cdot (\frac {x}{-y})^{3}÷(\frac {x^{2}y}{x})^{2};$
(11)$(-\frac {x}{y})^{2}\cdot (-\frac {y}{x})^{3}÷(\frac {1}{xy})^{2};$ (12)$(\frac {a}{b-a})^{4}÷\frac {1}{a^{2}-b^{2}}\cdot (\frac {a-b}{ab})^{3}.$
答案:(1)解:原式$=\frac{(-2a^{2}b)^{2}}{(3c)^{2}}=\frac{4a^{4}b^{2}}{9c^{2}}$
(2)解:原式$=\frac{(-3x^{2}y)^{3}}{(2x)^{3}}=\frac{-27x^{6}y^{3}}{8x^{3}}=-\frac{27x^{3}y^{3}}{8}$
(3)解:原式$=\frac{4x}{3y}\cdot\frac{y}{2x^{4}}\cdot x^{2}=\frac{4x\cdot y\cdot x^{2}}{3y\cdot2x^{4}}=\frac{4x^{3}y}{6x^{4}y}=\frac{2}{3x}$
(4)解:原式$=\frac{a^{2}}{b^{2}}\cdot\frac{4b}{3a}\cdot\frac{2b}{3a}=\frac{a^{2}\cdot4b\cdot2b}{b^{2}\cdot3a\cdot3a}=\frac{8a^{2}b^{2}}{9a^{2}b^{2}}=\frac{8}{9}$
(5)解:原式$=\frac{9a^{2}c^{2}}{4b^{2}}\cdot(-\frac{1}{9ac^{2}})=-\frac{9a^{2}c^{2}}{4b^{2}\cdot9ac^{2}}=-\frac{a}{4b^{2}}$
(6)解:原式$=\frac{-a^{6}b^{3}}{c^{3}}\cdot\frac{c^{4}}{a^{2}b^{2}}\cdot\frac{a^{4}}{b^{4}c^{4}}=-\frac{a^{6}b^{3}\cdot c^{4}\cdot a^{4}}{c^{3}\cdot a^{2}b^{2}\cdot b^{4}c^{4}}=-\frac{a^{8}}{b^{3}c^{3}}$
(7)解:原式$=\frac{a^{2}}{b^{2}}\cdot(-\frac{a^{3}}{b^{3}})\cdot(-\frac{1}{a^{4}b})=\frac{a^{2}\cdot a^{3}}{b^{2}\cdot b^{3}\cdot a^{4}b}=\frac{a^{5}}{a^{4}b^{6}}=\frac{a}{b^{5}}$
(8)解:原式$=\frac{4a^{2}b^{6}}{c^{4}d^{2}}\cdot\frac{b^{3}}{6a^{2}}\cdot(-\frac{27c^{3}}{b^{6}})=-\frac{4a^{2}b^{6}\cdot b^{3}\cdot27c^{3}}{c^{4}d^{2}\cdot6a^{2}\cdot b^{6}}=-\frac{18b^{2}}{cd^{2}}$
(9)解:原式$=6xy^{2}÷(-\frac{27y^{6}}{8x^{3}})\cdot\frac{4y^{2}}{9x^{2}}=6xy^{2}\cdot(-\frac{8x^{3}}{27y^{6}})\cdot\frac{4y^{2}}{9x^{2}}=-\frac{6xy^{2}\cdot8x^{3}\cdot4y^{2}}{27y^{6}\cdot9x^{2}}=-\frac{64x^{2}}{81y^{2}}$
(10)解:原式$=\frac{y^{6}}{x^{2}}\cdot(-\frac{x^{3}}{y^{3}})÷ x^{2}y^{2}=\frac{y^{6}}{x^{2}}\cdot(-\frac{x^{3}}{y^{3}})\cdot\frac{1}{x^{2}y^{2}}=-\frac{y^{6}\cdot x^{3}}{x^{2}\cdot y^{3}\cdot x^{2}y^{2}}=-\frac{y}{x}$
(11)解:原式$=\frac{x^{2}}{y^{2}}\cdot(-\frac{y^{3}}{x^{3}})\cdot x^{2}y^{2}=-\frac{x^{2}\cdot y^{3}\cdot x^{2}y^{2}}{y^{2}\cdot x^{3}}=-xy^{3}$
(12)解:原式$=\frac{a^{4}}{(b-a)^{4}}\cdot(a^{2}-b^{2})\cdot\frac{(a-b)^{3}}{a^{3}b^{3}}=\frac{a^{4}(a+b)(a-b)\cdot(a-b)^{3}}{(a-b)^{4}\cdot a^{3}b^{3}}=\frac{a(a+b)}{b^{3}}=\frac{a^{2}+ab}{b^{3}}$
解析:
 
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