【典型例题 1】计算：
(1)$8x^{2}y^{4}\cdot (-\frac {3x}{4y^{3}})÷(-\frac {x^{2}y}{2})$；
(2)$-\frac {x-y}{x+2y}÷\frac {x^{2}-y^{2}}{x^{2}+4xy+4y^{2}}\cdot \frac {x-y}{x+2y}$.

思路导引 分式的乘除混合运算,可以按照从左到右的顺序进行,也可以统一成乘法运算.
【解】(1)$8x^{2}y^{4}\cdot (-\frac {3x}{4y^{3}})÷(-\frac {x^{2}y}{2})$
$=(8x^{2}y^{4})\cdot (-\frac {3x}{4y^{3}})\cdot (-\frac {2}{x^{2}y})$
$=12x$.
(2)$-\frac {x-y}{x+2y}÷\frac {x^{2}-y^{2}}{x^{2}+4xy+4y^{2}}\cdot \frac {x-y}{x+2y}$
$=-\frac {x-y}{x+2y}\cdot \frac {(x+2y)^{2}}{(x+y)(x-y)}\cdot \frac {x-y}{x+2y}$
$=-\frac {x-y}{x+y}$.
规律方法 在进行分式的乘除混合运算时,应将分式的乘除混合运算统一化成乘法运算,这样可以运用乘法交换律和乘法结合律,从而使运算简便.若有多项式参与运算,应先将多项式分解因式,再约分,从而简化运算.

(1)
$8x^{2}y^{4}\cdot (-\frac {3x}{4y^{3}})÷(-\frac {x^{2}y}{2})$
$=(8x^{2}y^{4})\cdot (-\frac {3x}{4y^{3}})\cdot (-\frac {2}{x^{2}y})$
$=[8×(-\frac{3}{4})×(-2)]×(x^{2}\cdot x\cdot\frac{1}{x^{2}})×(y^{4}\cdot\frac{1}{y^{3}}\cdot\frac{1}{y})$
$ = 12x$
(2)
$-\frac {x - y}{x + 2y}÷\frac {x^{2}-y^{2}}{x^{2}+4xy + 4y^{2}}\cdot\frac {x - y}{x + 2y}$
$=-\frac {x - y}{x + 2y}\cdot\frac {(x + 2y)^{2}}{(x + y)(x - y)}\cdot\frac {x - y}{x + 2y}$
$=-\frac{x - y}{x + y}$

1. 计算：$\frac {m}{3m+9}\cdot \frac {6}{9-m^{2}}÷\frac {2m}{m-3}=$()
A.$\frac {1}{(m+3)^{2}}$
B.$-\frac {1}{(m+3)^{2}}$
C.$\frac {1}{(m-3)^{2}}$
D.$-\frac {1}{m^{2}+9}$

B

首先将分式乘除混合运算转化为乘法运算：
$\frac {m}{3m+9}\cdot \frac {6}{9-m^{2}}÷\frac {2m}{m-3}$
$=\frac {m}{3(m+3)}\cdot \frac {6}{(3+m)(3 - m)}×\frac {m-3}{2m}$
对$\frac {m}{3(m+3)}\cdot \frac {6}{(3+m)(3 - m)}×\frac {m-3}{2m}$进行化简，$m\neq0$，$m\neq\pm3$，
$=\frac{m}{3(m + 3)}×\frac{6}{(m + 3)(3 - m)}×\frac{m - 3}{2m}$
$=\frac{1}{(m+3)}×\frac{2}{(m + 3)(3 - m)}×(m - 3)×\frac{1}{2}$
因为$(3 - m)×(m - 3)=-(m - 3)^2$
$=\frac{1}{(m + 3)}×\frac{2×[-(m - 3)^2]}{2(m + 3)}$
$=-\frac{1}{(m + 3)^2}×\frac{(m - 3)^2}{(m - 3)^2}$
$=-\frac{1}{(m + 3)^2}$

2. 计算$\frac {a^{2}x}{b^{2}y}÷\frac {bx}{ay}\cdot \frac {by}{ax}$的结果为.

【解析】：原式$=\frac{a^{2}x}{b^{2}y}\cdot \frac{ay}{bx}\cdot \frac{by}{ax}=\frac{a^{2}x\cdot ay\cdot by}{b^{2}y\cdot bx\cdot ax}=\frac{a^{3}bxy^{2}}{ab^{3}x^{2}y}=\frac{a^{2}}{b^{2}x}$
【答案】：$\frac{a^{2}}{b^{2}x}$

【典型例题 2】计算：
(1)$(\frac {-nc^{3}}{3m})^{2}$；
(2)$(\frac {c}{ab})^{2}\cdot (-\frac {b}{ac})^{3}÷(-\frac {b}{a})^{4}$.
【解】(1)$(\frac {-nc^{3}}{3m})^{2}= \frac {(-nc^{3})^{2}}{(3m)^{2}}= \frac {n^{2}c^{6}}{9m^{2}}$.
(2)$(\frac {c}{ab})^{2}\cdot (-\frac {b}{ac})^{3}÷(-\frac {b}{a})^{4}= $
$-\frac {c^{2}}{a^{2}b^{2}}\cdot \frac {b^{3}}{a^{3}c^{3}}\cdot \frac {a^{4}}{b^{4}}= -\frac {1}{ab^{3}c}$.
规律方法 1. 进行分式的乘方运算时,一般先确定乘方结果的符号,正数的任何次方都是正数；负数的偶次方为正数,负数的奇次方为负数.
2. 分式的分子或分母是多项式时,应把多项式看作一个整体乘方.

(1)
$(\frac{-nc^{3}}{3m})^{2}$
$=\frac{(-nc^{3})^{2}}{(3m)^{2}}$
$=\frac{n^{2}c^{6}}{9m^{2}}$
(2)
$(\frac{c}{ab})^{2}\cdot(-\frac{b}{ac})^{3}÷(-\frac{b}{a})^{4}$
$=\frac{c^{2}}{a^{2}b^{2}}\cdot(-\frac{b^{3}}{a^{3}c^{3}})\cdot\frac{a^{4}}{b^{4}}$
$=-\frac{c^{2}}{a^{2}b^{2}}\cdot\frac{b^{3}}{a^{3}c^{3}}\cdot\frac{a^{4}}{b^{4}}$
$=-\frac{1}{ab^{3}c}$

3. 计算：
(1)$(-\frac {n}{m})^{3}\cdot (-\frac {m}{n})^{2}$；
(2)$(\frac {2x}{3y})^{2}\cdot (-\frac {3y}{4x})^{3}÷(\frac {1}{4}xy)$.

(1) $(-\frac{n}{m})^{3} \cdot (-\frac{m}{n})^{2}$
$=-\frac{n^{3}}{m^{3}} \cdot \frac{m^{2}}{n^{2}}$
$=-\frac{n^{3} \cdot m^{2}}{m^{3} \cdot n^{2}}$
$=-\frac{n}{m}$
(2) $(\frac{2x}{3y})^{2} \cdot (-\frac{3y}{4x})^{3} ÷ (\frac{1}{4}xy)$
$=\frac{4x^{2}}{9y^{2}} \cdot (-\frac{27y^{3}}{64x^{3}}) ÷ (\frac{1}{4}xy)$
$=\frac{4x^{2}}{9y^{2}} \cdot (-\frac{27y^{3}}{64x^{3}}) \cdot \frac{4}{xy}$
$=-\frac{4x^{2} \cdot 27y^{3} \cdot 4}{9y^{2} \cdot 64x^{3} \cdot xy}$
$=-\frac{3y^{2}}{4}$