已知(x+y)^2=8,(x-y)^2=5,求4xy与x^2+y^2的值

2025-03-10 04:39:08
推荐回答(5个)
回答1:

(x+y)²=8 x²+2xy+y²=8
(x-y)²=5 x²-2xy+y²=5
两式相加得:2x²+2y²=13
x²+y²=13/2
两式相减得:4xy=3

回答2:

已知(x+y)^2=8,(x-y)^2=5
4xy=(x+y)^2-(x-y)^2=3
x^2+y^2=(x+y)^2+(x-y)^2=13

回答3:

(x+y)^2=x^2+2xy+y^2=8,
(x-y)^2=x^2-2xy+y^2=5
二式相减得:4xy=3
二式相加得:x^2+y^2=(8+5)/2=13/2

回答4:

两式相减得
4xy=3
2xy=3/2
所以
x²+2xy+y²=8
x²+y²=8-2xy=8-3/2=13/2

回答5:

4xy=3 x^2+y^2=6.5