Question

求高手指点！

Answer 1

：因 |a-b|=b/a<1 => |a-b|<1 => -1<a-b<1 => a-b-1<0 a-b+1>0
(1/a-1/b)*)√（a-b-1)=(1/a-1/b)*|a-b-1| [ 因 a-b-1<0 => |a-b-1|=-(a-b-1)=b+1-a]
所原式=（1/a-1/b)*(b-a+1)
分类讨论（1）：当 a-b>0 |a-b|=a-b=b/a => b=a^2/(a+1) => 1/b=(a+1)/a^2
原式=[1/a-(a+1)/a^2]*[a^2/(a+1)-a+1]
=[a/a^2-(a+1)/a^2]*[(a^2+1-a^2)/(a+1)]
=[-1/a^2]*[1/(a+1)]
(2)当 a<b |a-b|=b-a=b/a b=a^2/(a-1) 1/b=(a-1)/a^2
原式=[1/a-(a-1)/a^2]*[a^2/(a-1)+1-a]
=[a/a^2-(a-1)/a^2]*[(2a-1)/(a-1)]
=(2a-1)/a^2*(a-1)