2+2

分析:把两式相减,运用分解因式法求x+y的值;把①×x-②×y可求xy的值.
把所求式子变形成含有x+y、xy的形式计算求解.
解答:两式相减,得
(x
2-y
2)+

(y-x)=0,
(x+y)(x-y)-

(x-y)=0,
(x-y)(x+y-

)=0,
∵x≠y,
∴x-y≠0,
∴x+y=

,
x
2+

y=

①,y
2+

x=

②,
①×x-②×y得 x
3-y
3=

(x-y),
∴x
2+xy+y
2=

,
(x+y)
2-xy=

,
∴xy=2-

.

+

=

=

=

,
=

-2=2(2+

)-2,
=2+2

.
点评:此题考查乘法公式的变形运用,难度较大.