2018 AMC 12A 第 10 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。

10.

有多少个实数有序对 (x,y)(x, y) 满足下面的方程组?

x+3y=3 x + 3y = 3 xy=1 \big|\,|x| - |y|\,\big| = 1

How many ordered pairs of real numbers (x,y)(x, y) satisfy the following system of equations?

x+3y=3 x + 3y = 3 xy=1 \big|\,|x| - |y|\,\big| = 1

11

22

33

44

88

答案:C
知识点:绝对值方程组分类讨论
难度评级:1560
解答:

第二个方程给出 xy=±1|x| - |y| = \pm 1, 等价于 x=±y±1x = \pm y \pm 1。 代入 x+3y=3x + 3y = 3

x=y+1x = y + 1(x,y)=(32,12)(x, y) = \left(\tfrac32, \tfrac12\right)。 若 x=y1x = y - 1(x,y)=(0,1)(x, y) = (0, 1)。 若 x=y+1x = -y + 1, 又得到 (x,y)=(0,1)(x, y) = (0, 1)。 若 x=y1x = -y - 1(x,y)=(3,2)(x, y) = (-3, 2)

不同的解是 (3,2)(-3, 2)(0,1)(0, 1), 和 (32,12)\left(\tfrac32, \tfrac12\right), 它们都满足原方程,所以共有 33 个。

所以正确答案是 C

The second equation gives xy=±1,|x| - |y| = \pm 1, equivalently x=±y±1.x = \pm y \pm 1. Substituting into x+3y=3:x + 3y = 3:

If x=y+1,x = y + 1, then (x,y)=(32,12).(x, y) = \left(\tfrac32, \tfrac12\right). If x=y1,x = y - 1, then (x,y)=(0,1).(x, y) = (0, 1). If x=y+1,x = -y + 1, then again (x,y)=(0,1).(x, y) = (0, 1). If x=y1,x = -y - 1, then (x,y)=(3,2).(x, y) = (-3, 2).

The distinct solutions are (3,2),(-3, 2), (0,1),(0, 1), and (32,12),\left(\tfrac32, \tfrac12\right), all of which check, so there are 3.3.

Thus, the correct answer is C.

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