已知(x²+y²)(x²-1+y²)=12 ,求x²+y²的值
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已知(x²+y²)(x²-1+y²)=12 ,求x²+y²的值
已知(x²+y²)(x²-1+y²)=12 ,求x²+y²的值
已知(x²+y²)(x²-1+y²)=12 ,求x²+y²的值
(x²+y²)(x²-1+y²)=12 ,
(x²+y²)²-(x²+y²)-12=0
(x²+y²-4)(x²+y²+3)=0
x²+y²+3>0
x²+y²-4=0
x²+y²=4
令a=x²+y²
则a(a+1)=12
a²+a-12=0
(a+4)(a-3)=0
因为x²+y²≥0
所以x²+y²=3
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