计算 (根号下x+xy-根号下xy)(根号下y+y的平方+y)(x>0 y>0) 快啊.

问题描述:

计算 (根号下x+xy-根号下xy)(根号下y+y的平方+y)(x>0 y>0) 快啊.

[√(x+xy)-√(xy)][√(y+y²)+y]
=√x[√(1+y)-√y]√y[√(1+y)+√y]
=√(xy)(1+y-y)
=√(xy)