[4(xy-1)^2-(xy+2)(2-xy)]÷¼xy
=[4x^2y^2-8xy+4-(4-x^2y^2)]÷¼xy
=(4x^2y^2-8xy+4-4+x^2y^2)÷¼xy
=(5x^2y^2-8xy)÷¼xy
=20xy-32
当x=-2,y=5分之1时,
原式=20×(-2)×5分之1-32
=-8-32
=-40
解:原式=[4(xy-1)�0�5-(xy+2)(2-xy)]÷(xy/4)=(4x�0�5y�0�5-8xy+4-4+x�0�5y�0�5)×(4/xy)
=(5x�0�5y�0�5-8xy)×(4/xy)
=20xy-32又x=-2,y=1/5所以xy=-2/5,得:20xy-32=-8-32=-40。