已知x^2-4x+4y^2-4y+5=0,求（x^2+y^2)(x^2-y^2)-(x+y)^2(x-y)^2的值?

两式子化简得 X^2-4x+4+4y^2-4y+1=0
即：（x-2)^2+(2y-1)^2=0

可以看出：（x-2)^2=0 （2y-1)^2=0
x=2 y=1/2
再将后面式子化简 把x y 值代入即可

由x^2-4x+4y^2-4y+5=0得
（x-2)^2+(2y-1)^2=0
所以x=2，y=1/2;
代入（x^2+y^2)(x^2-y^2)-(x+y)^2(x-y)^2得
原式=15/8;

x^2-4x+4y^2-4y+5=0可化简为
(x-2)^2+(2y-1)^2=0
得：x=2
y=1/2
（x^2+y^2)(x^2-y^2)-(x+y)^2(x-y)^2
=2x^2y^2-2y^4
=2y^2(x^2-y^2)
代入x及y的值得
2*(1/2)^2[2^2-(1/2)^2]=15/8

x^2-4x+4y^2-4y+5=0
(x-2)^2+(2y-1)^2=0
x-2=0,x=2
2y-1=0.y=1/2
（x^2+y^2)(x^2-y^2)-(x+y)^2(x-y)^2
=x^4-y^4-(x^2-y^2)^2
=16-1/16-(4-1/4)^2
=255/16-9/16
=246/16
=123/8

x^2-4x+4y^2-4y+5=0(x^2-4x+4)+(4y^2-4y+1)=0(x-2)^2+(2y-1)^2=0x-2=02y-1=0x=2y=1/2原式=x^4-y^4-[(x+y)(x-y)]^2=x^4-y^4-(x^2-y^2)^2=x^4-y^4-x^4+2x^2y^2-y^4=2x^2y^2-2y^4=2*4*1/4-2*1/16=15/8