来自刘宏志的问题
方程组{x^2+xy-2y^2=0{x^2+4y=-2解的个数是()a.1b.2c.3d.4
方程组{x^2+xy-2y^2=0
{x^2+4y=-2解的个数是()
a.1b.2c.3d.4
1回答
2020-05-12 22:11
方程组{x^2+xy-2y^2=0{x^2+4y=-2解的个数是()a.1b.2c.3d.4
方程组{x^2+xy-2y^2=0
{x^2+4y=-2解的个数是()
a.1b.2c.3d.4
x^2+xy-2y^2=0x^2+xy+1/4y^2=9/4y^2(x+y/2)^2=(3y/2)^2x+y/2=±3y/2=>x=y或-2y1.当x=y时x^2+4y=-2=>y^2+4y+2=0△=16-8=8>0,有两个根2.当x=-2y时x^2+4y=-2=>4y^2+4y+2=0△=16-32=-16