1. 分式的乘法法则:
分式乘分式,用分子的积作为积的分子,分母的积作为积的分母
.
用式子表示为:$\frac {a}{b}\cdot \frac {c}{d}= $
$\frac {a}{b}\cdot \frac {c}{d}$
.
答案：分式乘分式,用分子的积作为积的分子,分母的积作为积的分母$\frac {a}{b}\cdot \frac {c}{d}$
2. 分式的除法法则:
分式除以分式,把除式的分子、分母颠倒位置后,与被除式相乘
.
用式子表示为:$\frac {a}{b}÷\frac {c}{d}= $
$\frac {a}{b}\cdot \frac {d}{c}$
.
答案：分式除以分式,把除式的分子、分母颠倒位置后,与被除式相乘$\frac {a}{b}\cdot \frac {d}{c}$
1. 计算$(-\frac {1}{2}x)^{2}\cdot \frac {4}{x}$的结果是 (
B
)
A.-x
B.x
C.2x
D.$x^{3}$
答案：B
解析：
$(-\frac{1}{2}x)^{2} \cdot \frac{4}{x}$
$=\frac{1}{4}x^{2} \cdot \frac{4}{x}$
$=x$
B
2. 化简$\frac {1}{x-1}÷\frac {1}{x^{2}-1}$的结果是 (
C
)
A.$\frac {1}{x+1}$
B.$x-1$
C.$x+1$
D.x
答案：C
解析：
解：$\frac{1}{x-1} ÷ \frac{1}{x^2 - 1}$
$=\frac{1}{x-1} × (x^2 - 1)$
$=\frac{1}{x-1} × (x-1)(x+1)$
$=x + 1$
C
3. 化简$\frac {a+1}{a^{2}-a}÷\frac {a+1}{a^{2}-2a+1}$的结果是 (
D
)
A.$\frac {a+1}{a}$
B.$\frac {a}{a-1}$
C.$\frac {1}{a-1}$
D.$\frac {a-1}{a}$
答案：D
解析：
解：原式$=\frac{a+1}{a(a-1)} × \frac{(a-1)^2}{a+1}$
$=\frac{a-1}{a}$
答案：D
4. 计算:
(1)$\frac {x}{y^{2}}÷\frac {x^{2}}{y^{3}}$;
(2)$(-\frac {n}{m})÷\frac {n}{m^{2}+m}$;
(3)$\frac {x^{2}-1}{2x}\cdot \frac {4x^{2}}{x+1}$;
(4)$\frac {x-y}{y}\cdot \frac {y}{x(x-y)}$;
(5)$\frac {x-y}{y+x}\cdot \frac {x+y}{x(y-x)}$;
(6)$\frac {1}{a-2}÷\frac {a}{a^{2}-4}$;
(7)$\frac {8x^{2}}{x^{2}-2x+1}÷\frac {4x}{x-1}$;
(8)$\frac {9y^{2}-x^{2}}{x^{2}-2xy+y^{2}}÷\frac {x-3y}{2x^{2}-2xy}$.
答案：解:(1)$\frac {x}{y^{2}}÷\frac {x^{2}}{y^{3}}=\frac {x}{y^{2}}\cdot \frac {y^{3}}{x^{2}}=\frac {y}{x}.$
(2)$(-\frac {n}{m})÷\frac {n}{m^{2}+m}=-\frac {n}{m}\cdot \frac {m^{2}+m}{n}=-m-1.$
(3)$\frac {x^{2}-1}{2x}\cdot \frac {4x^{2}}{x+1}=\frac {(x+1)(x-1)}{2x}\cdot \frac {4x^{2}}{x+1}=2x^{2}-2x.$
(4)$\frac {x-y}{y}\cdot \frac {y}{x(x-y)}=\frac {1}{x}.$
(5)$\frac {x-y}{y+x}\cdot \frac {x+y}{x(y-x)}=\frac {x-y}{x+y}\cdot \frac {x+y}{-x(x-y)}=-\frac {1}{x}.$
(6)$\frac {1}{a-2}÷\frac {a}{a^{2}-4}=\frac {1}{a-2}\cdot \frac {(a+2)(a-2)}{a}=\frac {a+2}{a}.$
(7)$\frac {8x^{2}}{x^{2}-2x+1}÷\frac {4x}{x-1}=\frac {8x^{2}}{(x-1)^{2}}\cdot \frac {x-1}{4x}=\frac {2x}{x-1}.$
(8)$\frac {9y^{2}-x^{2}}{x^{2}-2xy+y^{2}}÷\frac {x-3y}{2x^{2}-2xy}=\frac {(3y+x)(3y-x)}{(x-y)^{2}}\cdot \frac {2x(x-y)}{x-3y}=\frac {-6xy+2x^{2}}{x-y}.$
