8年级数学题求解答已知x²+4y²
已知x²+4y²-4x+4y+5=0,求[（x^4-y^4）\xy]*[(x²-xy)\(xy-y²)]\[(x²+y²)\y]²的值
(1) x²+4y²-4x+4y+5=0=(x-2)²+4(y+1/2)²=0,x=2,y=-1/2 (2）代值x=2,y=-1/2，可得 [（x^4-y^4）\xy]*[(x²-xy)\(xy-y²)]\[(x²+y²)\y]²的值