若(x^2+y^2)(x^2+y^2-1)-12=0,且xy=5/2,求x+y的值

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若(x^2+y^2)(x^2+y^2-1)-12=0,且xy=5/2,求x+y的值

若(x^2+y^2)(x^2+y^2-1)-12=0,且xy=5/2,求x+y的值
若(x^2+y^2)(x^2+y^2-1)-12=0,且xy=5/2,求x+y的值

若(x^2+y^2)(x^2+y^2-1)-12=0,且xy=5/2,求x+y的值
设a=x²+y²,则
(x²+y²)(x²+y²-1)-12=0
a(a-1)-12=0
a²-a-12=0
(a-4)(a+3)=0
∴a=4或a=-3
∵a=x²+y²≥0,
∴a=-3不合题意,舍去
∴a=x²+y²=4,
x²+y²+2xy=(x+y)²=4+2×(5/2)=4+5=9
∴x+y=±√9=±3

令x²+y²=m>0
则由题意知m(m-1)-12=0
m²-m-12=0
(m-4)(m+3)=0
因为m>0
所以m=4
所以x²+y²=4
而xy=5/2
所以(x+y)²=x²+2xy+y²=4+5=9
所以x+y=±3

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