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典例1
解 $\dfrac{2x^2 + 2}{x^2 - 2x + 1} ÷ (x - 1 + \dfrac{x^2 + 2x + 1}{x - 1})$
$= \dfrac{2(x^2 + 1)}{(x - 1)^2} ÷ ( \dfrac{x^2 - 2x + 1}{x - 1} + \dfrac{x^2 + 2x + 1}{x - 1} ) = \dfrac{2(x^2 + 1)}{(x - 1)^2} · \dfrac{x - 1}{2(x^2 + 1)}$
$= \dfrac{1}{x - 1}.$
典例2
解 (1)由题意,得甲工程队的工作效率为$\dfrac{1000}{x}$(m/天),乙工程队的工作效率为$\dfrac{1000}{y}$(m/天),所以,甲工程队与乙工程队的工作效率之差为$\dfrac{1000}{x} - \dfrac{1000}{y} = \dfrac{1000(y - x)}{xy}$(m/天).
(2)甲、乙两工程队合作共需要的天数为$1000 ÷ ( \dfrac{1000}{x} + \dfrac{1000}{y} ) = \dfrac{xy}{x + y}.$
解 $\dfrac{2x^2 + 2}{x^2 - 2x + 1} ÷ (x - 1 + \dfrac{x^2 + 2x + 1}{x - 1})$
$= \dfrac{2(x^2 + 1)}{(x - 1)^2} ÷ ( \dfrac{x^2 - 2x + 1}{x - 1} + \dfrac{x^2 + 2x + 1}{x - 1} ) = \dfrac{2(x^2 + 1)}{(x - 1)^2} · \dfrac{x - 1}{2(x^2 + 1)}$
$= \dfrac{1}{x - 1}.$
典例2
解 (1)由题意,得甲工程队的工作效率为$\dfrac{1000}{x}$(m/天),乙工程队的工作效率为$\dfrac{1000}{y}$(m/天),所以,甲工程队与乙工程队的工作效率之差为$\dfrac{1000}{x} - \dfrac{1000}{y} = \dfrac{1000(y - x)}{xy}$(m/天).
(2)甲、乙两工程队合作共需要的天数为$1000 ÷ ( \dfrac{1000}{x} + \dfrac{1000}{y} ) = \dfrac{xy}{x + y}.$
答案
1. 化简$(1-\dfrac{1}{m})·\dfrac{1}{m-1}$的结果是（
A.$\dfrac{1}{m-1}$
B.$m$
C.$m-1$
D.$\dfrac{1}{m}$
D
）.A.$\dfrac{1}{m-1}$
B.$m$
C.$m-1$
D.$\dfrac{1}{m}$
答案
1. D
2. 为了练习分式的化简，张老师让同学们在分式$\frac{a^2}{a-2}$和$\frac{4a-4}{2-a}$中间加上“+”“−”“×”“÷”四个运算符号中的任意一个后进行化简，若化简的结果为$a-2$，则所加的运算符号为(
A.+
B.−
C.×
D.÷
A
).A.+
B.−
C.×
D.÷
答案
2. A
3. 计算：$(\dfrac{1}{b})^{3}· (ab)^{2}-(\dfrac{a}{b})^{3}÷ \dfrac{a}{b^{2}}=\_\_\_\_\_\_.$
答案
3. 0
4. 计算:
(1) $\dfrac{a}{a+1} · \dfrac{a^2 -1}{a^2} + \dfrac{1}{a}$;
(2) $(x - \dfrac{2x -1}{x}) ÷ \dfrac{x^2 -1}{x}$.
(1) $\dfrac{a}{a+1} · \dfrac{a^2 -1}{a^2} + \dfrac{1}{a}$;
(2) $(x - \dfrac{2x -1}{x}) ÷ \dfrac{x^2 -1}{x}$.
答案
4. 解 (1)$\dfrac{a}{a+1} · \dfrac{a^2 -1}{a^2} + \dfrac{1}{a}$
$=\dfrac{a}{a+1} · \dfrac{(a+1)(a-1)}{a^2} + \dfrac{1}{a}$
$=\dfrac{a-1}{a}+\dfrac{1}{a}$
$=\dfrac{a}{a}$
$=1.$
(2) $(x - \dfrac{2x -1}{x}) ÷ \dfrac{x^2 -1}{x}$
$=\dfrac{x^2-2x+1}{x} · \dfrac{x}{x^2-1}$
$=\dfrac{(x-1)^2}{x} · \dfrac{x}{(x+1)(x-1)}$
$=\dfrac{x-1}{x+1}.$
$=\dfrac{a}{a+1} · \dfrac{(a+1)(a-1)}{a^2} + \dfrac{1}{a}$
$=\dfrac{a-1}{a}+\dfrac{1}{a}$
$=\dfrac{a}{a}$
$=1.$
(2) $(x - \dfrac{2x -1}{x}) ÷ \dfrac{x^2 -1}{x}$
$=\dfrac{x^2-2x+1}{x} · \dfrac{x}{x^2-1}$
$=\dfrac{(x-1)^2}{x} · \dfrac{x}{(x+1)(x-1)}$
$=\dfrac{x-1}{x+1}.$
