2008-I-01 Posted on 21-06-2023 By app.cch 在〈2008-I-01〉中尚無留言 答案:$ab^3$$\begin{array}{cl} & \dfrac{(ab)^3}{a^2} \\ = & \dfrac{a^3b^3}{a^2} \\ = & a^{3-2} b^3 \\ = & ab^3 \end{array}$ Same Topic: 2007-I-02 2008-II-01 2008-II-39 2009-I-02 2008, 卷一, 香港中學會考 Tags:指數定律