2016 AMC 12A 第 7 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。

7.

下列哪一项描述了 x2(x+y+1)=y2(x+y+1)x^2(x+y+1)=y^2(x+y+1) 的图像?

Which of these describes the graph of x2(x+y+1)=y2(x+y+1)?x^2(x+y+1)=y^2(x+y+1)?

两条平行直线

two parallel lines

两条相交直线

two intersecting lines

三条直线都经过同一个公共点

three lines that all pass through a common point

三条直线不全经过同一个公共点

three lines that do not all pass through a common point

一条直线和一条抛物线

a line and a parabola

答案:D
知识点:因式分解平方差
难度评级:1410
小提示:

把所有项移到一边,并提出 x+y+1x+y+1

Move everything to one side and factor out x+y+1x+y+1

大提示:

(xy)(x+y)(x+y+1)=0(x-y)(x+y)(x+y+1)=0 是三条直线的并。

(xy)(x+y)(x+y+1)=0(x-y)(x+y)(x+y+1)=0 is the union of three lines

解答:

将所有项移到一边,得到 (x2y2)(x+y+1)=0 (x^2-y^2)(x+y+1)=0\text{,} 再分解为 (xy)(x+y)(x+y+1)=0 (x-y)(x+y)(x+y+1)=0\text{。} 因此图像是直线 x=yx=yx=yx=-yx+y+1=0x+y+1=0 的并集。

前两条直线交于原点,但第三条直线 x+y=1x+y=-1x+y=0x+y=0 平行,不经过原点。因此,图像由三条不全经过同一个公共点的直线组成。

所以正确答案是 D

Moving all terms to one side gives (x2y2)(x+y+1)=0, (x^2-y^2)(x+y+1)=0, which factors as (xy)(x+y)(x+y+1)=0. (x-y)(x+y)(x+y+1)=0. The graph is therefore the union of the lines x=y,x=y, x=y,x=-y, and x+y+1=0.x+y+1=0.

The first two lines intersect at the origin, but the third line x+y=1x+y=-1 is parallel to x+y=0x+y=0 and does not pass through the origin. So the graph consists of three lines that do not all pass through a common point.

Thus, the correct answer is D.

第 6 题#6
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