Question

解绝对值不等式？

Answer 1

|(x-1)/(x+1)|＜1
==> -1＜(x-1)/(x+1)＜1
左边：==> (x-1)/(x+1)+1＞0
==> [(x-1)+(x+1)]/(x+1)＞0
==> 2x/(x+1)＞0
==> x＞0，或x＜-1……………………………………①
右边：==> (x-1)/(x+1)-1＜0
==> [(x-1)-(x+1)]/(x+1)＜0
==> -2/(x+1)＜0
==> 2/(x+1)＞0
==> x＞-1…………………………………………………②
综上①②得：x＞0
或者两边平方更简单，你可以自己试试看。

Answer 2

|(x一1)/(x+1)|<1
一1<(ⅹ一1)/(x+1)<1
∴(x一1)/(x+1)>一1……(1)
(ⅹ一1)/(x+1)<1……(2)
由(1)得:(x一1)/(x+1)+1>0
2x/(x+1)>0
∴x>0或x<一1，
由(2)得:(x一1)/(x+1)一1<0
一2/(x+1)<0
∴x>一1，
综合(1)(2)可得:
原绝对值不等式的解是:ⅹ>0。