已知(x+y-1)^2+(xy+2)^2=0,求x^4+y^4的值

由题，
x+y-1=0,
xy+2=0,

x+y=1,
xy=-2

x^4+y^4
=(x^2+y^2)^2-2x^2y^2
=(x^2+y^2)^2-8
=[(x+y)^2-2xy]^2-8
=(1+4)^2-8
=25-8
=17

(x+y-1)^2+(xy+2)^2=0

上式成立必须有x+y-1=0，y=1-x
xy+2=0

xy+2=x(1-x)+2=0
x^2-x-2=0
(x-2)(x+1)=0
x1=2,y1=1-x1=1-2=-1
x2=-1, y=1-x2=1-(-1)=2

因此x^4+y^4=2^4+(-1)^4=17
或x^4+y^4=(-1)^4+2^4=17

答：x^4+y^4=17

(x+y-1)^2+(xy+2)^2=0,非负数的平方相加等于0,则
x+y-1=0,
xy+2=0,

x+y=1.xy=-2

x^4+y^4
=(x^2+y^2)-2x^2y^2
=[(x+y)^2-2xy]^2-2(xy)^2
=(1+4)^2-2*4
=17

(x+y-1)^2+(xy+2)^2=0,
所以只能
x+y-1=0
xy+2=0
x=2 y=-1或x=-1 y=2
x^4+y^4=17