2004 AMC 10A 第 2 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

2.

对任意三个实数 aa、bb、cc,其中 b≠cb \neq c,定义运算 ⋄\diamond 为 ⋄(a,b,c)=ab−c。\diamond(a, b, c) = \dfrac{a}{b - c}\text{。}求 ⋄(⋄(1,2,3),⋄(2,3,1),⋄(3,1,2))\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2)) 的值。

For any three real numbers a,a, b,b, and c,c, with b≠c,b \neq c, the operation ⋄\diamond is defined by ⋄(a,b,c)=ab−c.\diamond(a, b, c) = \dfrac{a}{b - c}. What is ⋄(⋄(1,2,3),⋄(2,3,1),⋄(3,1,2))?\diamond(\diamond(1, 2, 3), \diamond(2, 3, 1), \diamond(3, 1, 2))?

−12-\dfrac{1}{2}

−14-\dfrac{1}{4}

00

14\dfrac{1}{4}

12\dfrac{1}{2}

答案:B
知识点:自定义运算分数
难度评级:980
小提示:

先计算三个内层表达式。

Evaluate the three inner expressions before combining them

大提示:

分别计算 ⋄(1,2,3)=−1\diamond(1,2,3) = -1、⋄(2,3,1)=1\diamond(2,3,1) = 1、⋄(3,1,2)=−3\diamond(3,1,2) = -3。

⋄(1,2,3)=−1,\diamond(1,2,3) = -1, ⋄(2,3,1)=1,\diamond(2,3,1) = 1, and ⋄(3,1,2)=−3\diamond(3,1,2) = -3

解答:

三个内层表达式的值为 ⋄(1,2,3)=12−3=−1,⋄(2,3,1)=23−1=1,⋄(3,1,2)=31−2=−3。 \begin{aligned} \diamond(1,2,3) &= \dfrac{1}{2-3} = -1, \\ \diamond(2,3,1) &= \dfrac{2}{3-1} = 1, \\ \diamond(3,1,2) &= \dfrac{3}{1-2} = -3 \end{aligned}\text{。}

因此 ⋄(−1,1,−3)=−11−(−3)=−14。 \begin{aligned} \diamond(-1, 1, -3) &= \dfrac{-1}{1 - (-3)} \\ &= -\dfrac{1}{4} \end{aligned}\text{。}

所以正确答案是 B。

The inner values are ⋄(1,2,3)=12−3=−1,⋄(2,3,1)=23−1=1,⋄(3,1,2)=31−2=−3. \begin{aligned} \diamond(1,2,3) &= \dfrac{1}{2-3} = -1, \\ \diamond(2,3,1) &= \dfrac{2}{3-1} = 1, \\ \diamond(3,1,2) &= \dfrac{3}{1-2} = -3. \end{aligned}

Therefore ⋄(−1,1,−3)=−11−(−3)=−14. \begin{aligned} \diamond(-1, 1, -3) &= \dfrac{-1}{1 - (-3)} \\ &= -\dfrac{1}{4}. \end{aligned}

Thus, the correct answer is B.

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