2011 AMC 10B 第 15 题

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15.

@@ 表示“与……取平均”的运算:a@b=a+b2a @ b = \frac{a+b}{2}。下列哪些分配律对所有数 x,yx, yzz 都成立?

I. x@(y+z)=(x@y)+(x@z)x @ (y + z) = (x @ y) + (x @ z)

II. x+(y@z)=(x+y)@(x+z)x + (y @ z) = (x + y) @ (x + z)

III. x@(y@z)=(x@y)@(x@z)x @ (y @ z) = (x @ y) @ (x @ z)

Let @@ denote the "averaged with" operation: a@b=a+b2.a @ b = \frac{a+b}{2}. Which of the following distributive laws hold for all numbers x,y,x, y, and z?z?

I. x@(y+z)=(x@y)+(x@z)x @ (y + z) = (x @ y) + (x @ z)

II. x+(y@z)=(x+y)@(x+z)x + (y @ z) = (x + y) @ (x + z)

III. x@(y@z)=(x@y)@(x@z)x @ (y @ z) = (x @ y) @ (x @ z)

I only

II only

III only

I and III only

II and III only

答案:E
知识点:自定义运算分配律
难度评级:1280
解答:

I 的左边为 x+y+z2\dfrac{x+y+z}2,右边为 x+y+z2x+\dfrac{y+z}2,不恒相等。

II 的左边为 x+y+z2x+\dfrac{y+z}2,右边也为 ,恒成立。

III 的左边为 ,右边也为 2x+y+z4\dfrac{2x+y+z}4,恒成立。

所以正确答案是 E

In statement I, the left-hand side equals x+y+z2\dfrac{x+y+z}2 and the right-hand side equals x+y+z2,x+\dfrac{y+z}2, so they are not always equal.

In statement II, both sides equal x+y+z2,x+\dfrac{y+z}2, so it holds.

In statement III, both sides equal 2x+y+z4,\dfrac{2x+y+z}4, so it also holds.

Thus, the correct answer is E .

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