Question

下列哪一项描述了 $x^2(x+y+1)=y^2(x+y+1)$ 的图像？

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

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

解答：

将所有项移到一边，得到再分解为因此图像是直线 $x=y$ 、 $x=-y$ 和 $x+y+1=0$ 的并集。 $(x^2-y^2)(x+y+1)=0,$ $(x-y)(x+y)(x+y+1)=0.$

前两条直线交于原点，但第三条直线 $x+y=-1$ 与 $x+y=0$ 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 $(x^2-y^2)(x+y+1)=0,$ which factors as $(x-y)(x+y)(x+y+1)=0.$ The graph is therefore the union of the lines $x=y,$ $x=-y,$ and $x+y+1=0.$

The first two lines intersect at the origin, but the third line $x+y=-1$ is parallel to $x+y=0$ 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.

2016 AMC 12A 第 7 题

答案：D

难度评级：1410

解答：

其他年份的第 7 题