初二数学http://zhidao.baidu.com/question/49878389.html这题的过程

x^2+4y^2+x^2y^2+1=6xy
x^2+4y^2+x^2y^2+1-6xy=0
x^2-4xy+4y^2+x^2y^2-2xy+1=0
(x-2y)^2+(xy-1)^2=0
(x-2y)^2=0,(xy-1)^2=0
x=2y,
xy=1
x=2y,
2y*y=1
y^2=1/2
y=±√2/2
(x^4-y^4)/(2x^2+xy-y^2)*(2x-y)/(xy-y^2)
=(x^2-y^2)(x^2+y^2)/[(2x-y)(x+y)]*(2x-y)/[y(x-y)]
=(x^2-y^2)(x^2+y^2)/(x+y)*1/[y(x-y)]
=(x-y)(x+y)(x^2+y^2)/(x+y)*1/[y(x-y)]
=y/(x^2+y^2)
=y/(x^2+2xy+y^2-2xy)
=y/[(x+y)^2-2xy]
=y/[(2y+y)^2-2*1]
=y/[(9y^2-2]
=(±√2/2)/[(9*1/2-2]
=(±√2/2)/(9/2-2)
=(±√2/2)/(5/2)
=±√2/5

做什么？