AMC10 2011 B

AMC10 2011 B · Q15

AMC10 2011 B · Q15. It mainly tests Manipulating equations, Algebra misc.

Let @ denote the “averaged with” operation: a @ b = $\frac{a+b}{2} $. Which of the following distributive laws hold for all numbers x, y, and z ? I. x @ (y + z) = (x @ y) + (x @ z) II. x + (y @ z) = (x + y) @ (x + z) III. x @ (y @ z) = (x @ y) @ (x @ z)

设 @ 表示“平均”运算：a @ b = $\frac{a+b}{2}$。以下哪些分配律对所有数 x, y, z 成立？ I. x @ (y + z) = (x @ y) + (x @ z) II. x + (y @ z) = (x + y) @ (x + z) III. x @ (y @ z) = (x @ y) @ (x @ z)

(A) I only 仅 I

(B) II only 仅 II

(D) I and III only I 和 III

(E) II and III only II 和 III

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): If $x\neq 0$, then I is false: $x\odot(y+z)=\dfrac{x+(y+z)}{2}\neq\dfrac{x+y+x+z}{2}=\dfrac{x+y}{2}+\dfrac{x+z}{2}=(x\odot y)+(x\odot z).$ On the other hand, II and III are true for all values of $x$, $y$ and $z$: $x+(y\odot z)=x+\dfrac{y+z}{2}=\dfrac{2x+y+z}{2}=\dfrac{(x+y)+(x+z)}{2}=(x+y)\odot(x+z),$ and $x\odot(y\odot z)=\dfrac{x+\dfrac{y+z}{2}}{2}=\dfrac{\left(\dfrac{2x+y+z}{2}\right)}{2}=\dfrac{\dfrac{x+y}{2}+\dfrac{x+z}{2}}{2}=(x\odot y)\odot(x\odot z)$

答案（E）：如果 $x\neq 0$，则命题 I 为假： $x\odot(y+z)=\dfrac{x+(y+z)}{2}\neq\dfrac{x+y+x+z}{2}=\dfrac{x+y}{2}+\dfrac{x+z}{2}=(x\odot y)+(x\odot z).$ 另一方面，对所有 $x$、$y$、$z$ 的取值，命题 II 和 III 都为真： $x+(y\odot z)=x+\dfrac{y+z}{2}=\dfrac{2x+y+z}{2}=\dfrac{(x+y)+(x+z)}{2}=(x+y)\odot(x+z),$ 并且 $x\odot(y\odot z)=\dfrac{x+\dfrac{y+z}{2}}{2}=\dfrac{\left(\dfrac{2x+y+z}{2}\right)}{2}=\dfrac{\dfrac{x+y}{2}+\dfrac{x+z}{2}}{2}=(x\odot y)\odot(x\odot z)$

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