已知(x+y)^2=8,(x-y)^2=4,求x^2+y^2及xy的值

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已知(x+y)^2=8,(x-y)^2=4,求x^2+y^2及xy的值
已知(x+y)^2=8,(x-y)^2=4,求x^2+y^2及xy的值

已知(x+y)^2=8,(x-y)^2=4,求x^2+y^2及xy的值
(x+y)^2+(x-y)^2
=x^2+2xy+y^2+x^2-2xy+y^2
=2(x^2+y^2)=8+4=12
x^2+y^2=6
(x+y)^2-(x-y)^2
=x^2+2xy+y^2-x^2+2xy-y^2
=4xy=8-4=4
xy=1

因为(x+y)^2=8, (x-y)^2=4
所以 (x-y)^2+4xy=(x+y)^2=8
故4xy=4 则xy=1
因此x^2+y^2=(x+y)^2-2xy=8-2=6

x^2+2xy+y^2=8, x^2-2xy+y^2=4-->xy=1,x^2+y^2=6

x^2+y^2=a xy=b
(x+y)^2=8,(x-y)^2=4即a+2b=8 a-2b=4
a=2 b=1
即x^2+y^2=2 xy=1

(x+y)^2=8 (x-y)^2=4
x^2+y^2=2xy=8……1式 x^2+y^2-2xy=4……2式
1式+2式：2（x^2+y^2）=12
x^2+y^2=6
1式-2式：4xy=4
xy=1

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