2016 AMC 12A 第 7 题

先试着解答 2016 AMC 12A 第 7 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2016 AMC 12A 解答,或核对答案。

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
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:
(x−y)(x+y)(x+y+1)=0(x-y)(x+y)(x+y+1)=0 是三条直线的并。
(x−y)(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 (x2−y2)(x+y+1)=0, (x^2-y^2)(x+y+1)=0\text{,} 再分解为 (x−y)(x+y)(x+y+1)=0。 (x-y)(x+y)(x+y+1)=0\text{。} 因此图像是直线 x=yx=y、x=−yx=-y 和 x+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=-1 与 x+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 (x2−y2)(x+y+1)=0, (x^2-y^2)(x+y+1)=0, which factors as (x−y)(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.
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