题目

已知x=1+根号2,y=1-根号2,求(x-y分之1-x+y分之1)÷x²-2xy+y²分之2y的值;

提问时间：2020-10-31

答案

x=1+根号2,y=1-根号2,
∴x+y=2
x-y=2√2
(x-y分之1-x+y分之1)÷x²-2xy+y²分之2y
=[1/(x-y)-1/(x+y)×(x-y)²/2y
=(x+y-x+y)/(x+y)(x-y)×(x-y)²/2y
=2y/(x+y)×(x-y)/2y
=(x-y)/(x+y)
=2√2/2
=√2