2000 AMC 8 第 17 题

先试着解答 2000 AMC 8 第 17 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2000 AMC 8 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

对所有非零数,运算 \otimes 定义为 ab=a2b.a\otimes b =\dfrac{a^{2}}{b}.

[(12)3][1(23)].[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)].

The operation \otimes is defined for all nonzero numbers by ab=a2b.a\otimes b =\dfrac{a^{2}}{b}.

Determine [(12)3][1(23)].[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)].

23-\dfrac{2}{3}

14-\dfrac{1}{4}

00

14\dfrac{1}{4}

23\dfrac{2}{3}

答案:A
知识点:自定义运算分数
难度评级:1410
解答:

计算如下: [(12)3][1(23)]=[1223][1223]=[123][143]=(1/2)2312(4/3)=141334=11234=23. \begin{gather*} [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] \\ = [\dfrac{1^2}{2} \otimes 3] - [1 \otimes \dfrac{2^2}{3}] \\ = [\dfrac{1}{2} \otimes 3] - [1 \otimes \dfrac{4}{3}] \\ = \dfrac{(1 / 2)^2}{3} - \dfrac{1^2}{(4 / 3)} \\ = \dfrac{1}{4} \cdot \dfrac{1}{3} - \dfrac{3}{4} \\ = \dfrac{1}{12} - \dfrac{3}{4} \\= -\dfrac{2}{3}. \end{gather*}

所以正确答案是 A

We can calculate it as follows. [(12)3][1(23)]=[1223][1223]=[123][143]=(1/2)2312(4/3)=141334=11234=23. \begin{gather*} [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] \\ = [\dfrac{1^2}{2} \otimes 3] - [1 \otimes \dfrac{2^2}{3}] \\ = [\dfrac{1}{2} \otimes 3] - [1 \otimes \dfrac{4}{3}] \\ = \dfrac{(1 / 2)^2}{3} - \dfrac{1^2}{(4 / 3)} \\ = \dfrac{1}{4} \cdot \dfrac{1}{3} - \dfrac{3}{4} \\ = \dfrac{1}{12} - \dfrac{3}{4} \\= -\dfrac{2}{3}. \end{gather*}

Thus, A is the correct answer.

← 第 16 题#16
完整试卷

其他年份的第 17 题

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8