若(x²+y²)(1-x²-y²)+6=0,则x²+y²=

若(x²+y²)(1-x²-y²)+6=0,则x²+y²=
若(x²+y²)(1-x²-y²)+6=0,则x²+y²=

若(x²+y²)(1-x²-y²)+6=0,则x²+y²=
(x²+y²)(1-x²-y²)+6=0
(x²+y²)【1-（x²+y²)】+6=0
(x²+y²)²- (x²+y²)-6=0
令 (x²+y²)=z≥0
原式化为：z²-z-6=0
解得：(x²+y²)=z=3

等于3 x2+y2=a a*（1-a）=-6

这个我还没学，我6年级