题目

x^2+4y^2-4x+4y+5=0,求[（x^4-y^4)/xy]*[(2x^2-xy)/(xy-y^2)]/{(x^2+y^2)/y}^2.

提问时间：2020-07-24

答案

x^2+4y^2-4x+4y+5=0,
x^2-4x+4+4y^2+4y+1=0,
(x-2)^2+(2y+1)^2=0
x=2,y=-1/2
[（x^4-y^4)/xy]*[(2x^2-xy)/(xy-y^2)]/{(x^2+y^2)/y}^2.
=[(x^2-y^2)(x^2+y^2)/xy]*[x(2x-y)/y(x-y)]/{(x^2+y^2)/y}^2.
=[(x^2-y^2)(x^2+y^2)/xy]*[x(2x-y)/y(x-y)]*{y/(x^2+y^2)}^2.
=[(x^2-y^2)(x^2+y^2)/xy]*[x(2x-y)/y(x-y)]*y^2/(x^2+y^2)^2.
=[(x^2-y^2)/xy]*[x(2x-y)/y(x-y)]*y^2/(x^2+y^2).
=(x^2-y^2)*(2x-y)/(x-y)*1/(x^2+y^2).
=(x-y)(x+y)*(2x-y)/(x-y)*1/(x^2+y^2).
=(x+y)*(2x-y)/(x^2+y^2).
=[2+(-1/2)]/[2^2+(-1/2)^2]
=(3/2)/(17/4)
=3/2*4/17
=6/17