先化简再求值（x-y/1-x+y/1）÷x平方+2xy+y平方/2y,其中x=根号3+2,y=根号3

分类: 作业答案 • 2021-12-20 04:45:02

问题描述：

答

（x-y/1-x+y/1）÷x平方+2xy+y平方/2y
=根号3+1

答

原式=[1/(x-y)-1/(x+y)]/[2y/(x+y)²]
=[(x+y-x+y)/(x+y)(x-y)]*(x+y)²/2y
=(x+y)/(x-y)
=(√3+2+√3)/(√3+2-√3)
=√3+1