Question

定义运算“减去倒数”为 $a\diamond b=a-\frac{1}{b}$ 。求 $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))\text{？}$

Consider the operation “minus the reciprocal of,” defined by $a\diamond b=a-\frac{1}{b}.$ What is $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))?$

$-\dfrac{7}{30}$

$-\dfrac{1}{6}$

$0$

$\dfrac{1}{6}$

$\dfrac{7}{30}$

答案：A

知识点：自定义运算分数

难度评级：960

小提示：

先计算 $1\diamond2$ 和 $2\diamond3$ 。

Compute $1\diamond2$ and $2\diamond3$ first

大提示：

记住 $a\diamond b=a-\frac1b$ 。

Remember $a\diamond b=a-\frac1b$

解答：

$\begin{aligned} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{aligned}$

所以正确答案是 A。

$\begin{aligned} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{aligned}$

Thus, the correct answer is A .

2015 AMC 10B 第 7 题

答案：A

难度评级：960

小提示：

大提示：

解答：

其他年份的第 7 题