计算:(1/x-y-1/x+y)/2y/x^2-2xy+y^2,x=1+根号2,y=1-根号2

来源:学生作业学帮网 编辑:学帮网 时间:2024/05/28 01:23:31

计算:(1/x-y-1/x+y)/2y/x^2-2xy+y^2,x=1+根号2,y=1-根号2

[1/(x-y)-1/(x+y)]/[2y/(x^2-2xy+y^2)]
={[(x+y)-(x-y)]/[(x+y)(x-y)]}*(x^2-2xy+y^2)/2y
={2y/[(x+y)(x-y)]}*{(x-y)^2/2y}
=(x-y)/(x+y)
x=1+根号2,y=1-根号2,代入
原式=【(1+根号2)-(1-根号2)】/【(1+根号2)+(1-根号2)】
=(2根号2)/2
=根号2