初中数学（香港课程）/多项式的运算/2

指数定律一：$b^{p}\cdot b^{q}=b^{p+q}$

探究活动
展开$b^{2}$和$b^{6}$ 从而化简$b^{2}\cdot b^{6}$ 以同样方式化简$c^{3}\cdot c^{4}$ 写出结论

我们得出了先加上指数，再与底乘方等同于先乘方得出结果，再乘

严谨证明

${\begin{aligned}b^{p}\times b^{q}&=\overbrace {b\times b\times \cdots \times b} ^{p}\times \overbrace {b\times b\times \cdots \times b} ^{q}\\&=\underbrace {b\times b\times \cdots \times b} _{p+q}\\&=b^{p+q}\end{aligned}}$

示范例子1 化简 $9d^{7}(-8d^{5})$ $w^{6}\cdot 4w\cdot w^{2}$ $7pq^{3}(-p^{8}q)$ 解： $9d^{7}(-8d^{5})=9(-8)d^{7+5}=-72d^{1}2$ $w^{6}\cdot 4w\cdot w^{2}=4w^{9}$ $7pq^{3}(-p^{8}q)=-7p^{9}q^{4}$

同类习题1 {{{3}}}

指数定律二：${\frac {b^{p}}{b^{q}}}=b^{p-q}$

Extended content
展开$b^{3}$和$b^{8}$ 从而化简${\frac {b^{8}}{b^{3}}}$ 以同样方式化简${\frac {a^{9}}{a^{7}}}$

先将指数相减，再与底乘方等同于先乘方得出结果，再将结果相除

• 本课只会教授$p>q$的情况

严谨证明

${\begin{aligned}{\frac {b^{p}}{b^{q}}}&={\frac {b^{p-q+q}}{b^{q}}}\\&={\frac {b^{p-q}\cdot b^{q}}{b^{q}}}\\&=b^{p-q}\end{aligned}}$

示范例子1 化简 ${\frac {9x^{6}}{18x^{2}}}$ ${\frac {7u^{4}\cdot u^{5}}{u^{6}}}$ ${\frac {10a^{9}c^{8}}{2a^{7}c^{3}}}$ 解： 解析失败 (语法错误): {\displaystyle \frac{9x^6}{18x^2}=\frac{9}{18}x^{6-2}=\frac{x^4}{2} # <math>\frac{7u^4\cdot u^5}{u^6}=\frac{7u^9}{u^6}=7u^3} ${\frac {10a^{9}c^{8}}{2a^{7}c^{3}}}=5a^{2}c^{5}$

同类习题1 {{{3}}}

示范例子1 计算${\frac {6^{200}\cdot 6^{400}}{6^{598}}}$ ${\begin{aligned}\end{aligned}}$ ${\begin{aligned}{\frac {6^{200}\cdot 6^{400}}{6^{598}}}&={\frac {6^{600}}{6^{598}}}\\&=6^{2}=36\end{aligned}}$

同类习题1 {{{3}}}