若(x-y)^2=12,(x+y)^2=16,求xy的值

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若(x-y)^2=12,(x+y)^2=16,求xy的值

若(x-y)^2=12,(x+y)^2=16,求xy的值
若(x-y)^2=12,(x+y)^2=16,求xy的值

若(x-y)^2=12,(x+y)^2=16,求xy的值
(x+y)^2=16
(x-y)^2=12
相减得
4xy=4
xy=1

解: (x-y)²=12. ,(x+y)²=16
4xy=(x+y)²-(x-y)²=16-12=4
xy=4÷4=1
祝你学习进步,如有不明可以追问.同意我的答案请采纳,O(∩_∩)O谢谢

(x-y)^2=12,(x+y)^2=16
(x+y)^2 -(x-y)^2
=4xy=4
xy=1