若(x^2+y^2)(x^2+y^2-2)+3=0则x^2+y^2=?还有几道题帮下忙已知x^2+(k+1)+k^2+8=0有两个相等实数根，则k=关于x的方程kx^2-2x-1=0有实数根，则k方程mx^2-mx+1=0有两个相等实数根，则m=防侧滑那个（m+1)x^2+2x=1有两

来源：学生作业帮助网编辑：作业帮时间：2024/04/19 22:32:52

若(x^2+y^2)(x^2+y^2-2)+3=0则x^2+y^2=?还有几道题帮下忙已知x^2+(k+1)+k^2+8=0有两个相等实数根，则k=关于x的方程kx^2-2x-1=0有实数根，则k方程mx^2-mx+1=0有两个相等实数根，则m=防侧滑那个（m+1)x^2+2x=1有两
若(x^2+y^2)(x^2+y^2-2)+3=0则x^2+y^2=?
还有几道题帮下忙
已知x^2+(k+1)+k^2+8=0有两个相等实数根，则k=
关于x的方程kx^2-2x-1=0有实数根，则k
方程mx^2-mx+1=0有两个相等实数根，则m=
防侧滑那个（m+1)x^2+2x=1有两个不相等的实数根，则m

若(x^2+y^2)(x^2+y^2-2)+3=0则x^2+y^2=?还有几道题帮下忙已知x^2+(k+1)+k^2+8=0有两个相等实数根，则k=关于x的方程kx^2-2x-1=0有实数根，则k方程mx^2-mx+1=0有两个相等实数根，则m=防侧滑那个（m+1)x^2+2x=1有两
令t=x^2+y^2则t>=0
则原式=t(t-2)+3=0
无解
你的题目写错了吧

(x^2+y^2)(x^2+y^2-2)+3=0
(x^2+y^2)+[(x^2+y^2)-2]+3=0
(x^2+y^2)^2-2*(x^2+y^2)+3=0
[(x^2+y^2)-3][(x^2+y^2)+1]=0
x^2+y^2=3 x^2+y^2=-1(x^2+y^2>=0的不和题意，舍去)

设x^2+y^2＝m,则
m^2-2m+3=0
题目抄错了吧
应该是－3.
那样就可解得：m=3或m=-1( x^2+y^2不可能为负，舍去）

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