已知(x²+y²)²-y²=x²+6,求x²+y²的值
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已知(x²+y²)²-y²=x²+6,求x²+y²的值
已知(x²+y²)²-y²=x²+6,求x²+y²的值
已知(x²+y²)²-y²=x²+6,求x²+y²的值
(x²+y²)²-(x²+y²)-6=0
(x²+y²-3)(x²+y²+2)=0
x²+y²=3
(x²+y²)²=x²+y²+6
设x2+y2=t
则t2-t-6=0
分解因式(t-3)(t+2)=0
所以t=3或t=-2
因为x2+y2≥0
所以t取3
即x2+y2=3
望采纳。