【例1】计算$1÷(\dfrac{1}{x}+\dfrac{1}{y})$的结果是
$\dfrac{xy}{x+y}$
。答案
$\dfrac{xy}{x+y}$
练习1.计算:$\dfrac{x^2 - 4x + 4}{x + 2} ÷ \dfrac{x - 2}{x + 2}$的结果是
$x-2$
。答案
$x-2$
练习2.计算$\frac{y}{x^2 - y^2} ÷ (1 - \frac{x}{x + y})$的结果是
$\dfrac{1}{x-y}$
。答案
$\dfrac{1}{x-y}$
练习3.计算:
(1) $(1+\dfrac{1}{x})· \dfrac{x}{x^2 -1}$;
(2) $(a^2 +3a)÷ \dfrac{a^2 -9}{a -3}$;
(3) $(\dfrac{1}{a}+\dfrac{1}{b})^2 ÷ (\dfrac{1}{a^2}-\dfrac{1}{b^2})$;
(4) $(\dfrac{1}{x}+\dfrac{1}{y})÷ (\dfrac{1}{x}-\dfrac{1}{y})$;
(5) $(\dfrac{x}{x+1}-\dfrac{3x}{x-1})÷ \dfrac{x}{x^2 -1}$;
(6) $(1+\dfrac{2b}{a -b})^2 · (1-\dfrac{2b}{a +b})^2$。
(1) $(1+\dfrac{1}{x})· \dfrac{x}{x^2 -1}$;
(2) $(a^2 +3a)÷ \dfrac{a^2 -9}{a -3}$;
(3) $(\dfrac{1}{a}+\dfrac{1}{b})^2 ÷ (\dfrac{1}{a^2}-\dfrac{1}{b^2})$;
(4) $(\dfrac{1}{x}+\dfrac{1}{y})÷ (\dfrac{1}{x}-\dfrac{1}{y})$;
(5) $(\dfrac{x}{x+1}-\dfrac{3x}{x-1})÷ \dfrac{x}{x^2 -1}$;
(6) $(1+\dfrac{2b}{a -b})^2 · (1-\dfrac{2b}{a +b})^2$。
答案
解:(1)原式$=\dfrac{x+1}{x}·\dfrac{x}{(x+1)(x-1)}=\dfrac{1}{x-1}$;
(2)原式$=a(a+3)×\dfrac{a-3}{(a+3)(a-3)}=a$;
(3)原式$=(\dfrac{a+b}{ab})^2×\dfrac{a^2b^2}{b^2-a^2}=\dfrac{a+b}{b-a}$;
(4)原式$=\dfrac{y+x}{xy}×\dfrac{xy}{y-x}=\dfrac{x+y}{y-x}$;
(5)原式$=\dfrac{x^2-x-3x(x+1)}{(x+1)(x-1)}×\dfrac{(x+1)(x-1)}{x}$
$=-2x-4$;
(6)原式$=(\dfrac{a+b}{a-b})^2×\dfrac{(a-b)^2}{(a+b)^2}=1$。
(2)原式$=a(a+3)×\dfrac{a-3}{(a+3)(a-3)}=a$;
(3)原式$=(\dfrac{a+b}{ab})^2×\dfrac{a^2b^2}{b^2-a^2}=\dfrac{a+b}{b-a}$;
(4)原式$=\dfrac{y+x}{xy}×\dfrac{xy}{y-x}=\dfrac{x+y}{y-x}$;
(5)原式$=\dfrac{x^2-x-3x(x+1)}{(x+1)(x-1)}×\dfrac{(x+1)(x-1)}{x}$
$=-2x-4$;
(6)原式$=(\dfrac{a+b}{a-b})^2×\dfrac{(a-b)^2}{(a+b)^2}=1$。
【例2】采购员去粮油公司购买两次大米,两次大米的价格分别为m元/千克和n元/千克,每次去都用了y(元),则两次所购大米的平均价格是
$\dfrac{2mn}{m+n}$
元/千克.答案
$\dfrac{2mn}{m+n}$
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