已知(x+y)^2=4,(x-y)^2=1,求x^2+y^2及xy的值

x^2+2xy+y^2=4
x^2-2xy+y^2=1
两式相加2x^2+2y^2=5
x^2+y^2=5/2
两式相减4xy=3
xy=3/4

x^2+y^2=（(x+y)^2+（x-y)^2）/2=5/2
xy=（(x+y)^2-(x-y)^2)/4=3/4

(x+y)^2=x^2+2xy+y^2
(x-y)^2=x^2-2xy+y^2
(x+y)^2+(x-y)^2=2(x^2+y^2)
则x^2+y^2=5/2
(x+y)^2-(x-y)^2=4xy
则xy=3/4

(x+y)^2=X^2+Y^2+2XY
(x-y)^2=X^2+Y^2-2XY
x^2+y^2=[(X^2+Y^2+2XY)+(X^2+Y^2-2XY)]/2=5/2
xy=[(X^2+Y^2+2XY)-(X^2+Y^2-2XY)]/4=3/4