解:原式=[x(√x+√y)]/[y(x-y)]-[(√x)²+√x×√y+(√y)²]/[(√x)³-(√y)³]=[x(√x+√y)]/[y(√x+√y)(√x-√y)]-[(√x)²+√x×√y+(√y)²]/﹛(√x-√y)[(√x)²+√x×√y+(√y)²]﹜=x/[y(√x-√y)]-1/(√x-√y)=x/[y(√x-√y)]-y/[y(√x-√y)]=(x-y)/[y(√x-√y)]=[(√x+√y)(√x-√y)]/[y(√x-√y)]=(√x+√y)/y