初二数学计算题 已知2x-y=10,求（(x^2+y^2)-(x-y)^2+2y(x-y))/4y

（(x^2+y^2)-(x-y)^2+2y(x-y))/4y
=(x^2+y^2-(x^2-2xy+y^2)+(2xy-2y^2))/4y
=(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)/4y
=(4xy-2y^2)/4y
=2y(2x-y)/4y
=(2x-y)/2
当2x-y=10时
原式＝10/2＝5