已知x2 +y2=12,xy=3,且0＜x＜y,求（x-y)／(x+y)的值。

(x-y)²=x²-2xy+y²=6
x<y,所以x-y<0
所以x-y=-√6

(x+y)²=x²+2xy+y²=18
x>0,y>0y,所以x+y>0
所以x+y=√18

所以(x-y)/(x+y)=-√(6/18)=-√(1/3)=-√3/3

x^2+y^2+2xy=12+3*2=18
(x+y)^2=18
x^2+y^2-2xy=12-3*2=6
(x-y)^2=6
((x-y)/(x+y))^2=6/18=1/3
x-y<0,x+y>0
(x-y)/(x+y)<0
(x-y)／(x+y)=-1/3根号3

解：
∵x^2+y^2=12,xy=3
∴(x+y)^2=x^2+y^2+2xy=12+6=18
(x-y)^2=x^2+y^2-2xy=12-6=6
∵0＜x＜y
∴x-y＜0，x+y＞0
故可以得出：
x+y=3√2
x-y=-√6
∴原式=-√6/（3√3）=-√3/3

x^2+y^2=12;
xy=3;
x^2+2xy+y^2=(x+y)^2=12+2*3=18;
x+y=3√2；
x^2-2xy+y^2=(x-y)^2=12-2*3=6;
x-y=-√6；
（x-y)/(x+y)=(-√6)/(3√2)=-√3/3

x2+y2+2xy-2xy=12
(x+y)2=12+2xy
(x+y)2=18
(x-y)2=12-2xy
(x-y)2=6
(x-y)2/(x+y)2=6/18
x-y/x+y=√3/3