{（x-y)/(根号x+根号y)}-(x+y-2倍根号xy)/(根号x-根号y)=?

{（x-y)/(根号x+根号y)}-(x+y-2倍根号xy)/(根号x-根号y)=?
{（x-y)/(根号x+根号y)}-(x+y-2倍根号xy)/(根号x-根号y)=?

{（x-y)/(根号x+根号y)}-(x+y-2倍根号xy)/(根号x-根号y)=?
（（x-y)/(√x+√y))-(x+y-2√xy)/(√x-√y),分母有理化,第一个式子分母乘以√x-√y,又(x+y-2√xy)=(√x-√y)(√x-√y),所以原式等于√x-√y-（√x-√y）=0

（（x-y)/(√x+√y))-(x+y-2√xy)/(√x-√y),分母有理化，第一个式子分母乘以√x-√y，又(x+y-2√xy)=(√x-√y)(√x-√y),所以原式等于√x-√y-（√x-√y）=0