题是这样的吧：
[(√x-√y)^3+2x√x+y√y]/(x√x+y√y) + [3√(xy)-3y]/(x-y)
原式=[(x√x-3x√y+3y√x-y√y)+2x√x+y√y]/(x√x+y√y) + [3√(xy)-3y]/(x-y)
=(3x√x-3x√y+3y√x)]/(x√x+y√y) + [3√(xy)-3y]/(x-y)
=3√x[x-√(xy)+3y]/{(√x+√y)[x-√(xy)+3y]}+3√y(√x-√y)/[(√x+√y)(√x-√y)]
=(3√x)/(√x+√y) + (3√y)/(√x+√y)
=(3√x+ 3√y)/(√x+√y)
=3(√x+ √y)/(√x+√y)
=3
这题要用到三次方乘法公式和立方和公式