Question

求不等式|(x/x+1)-2|>3的解集

Answer 1

因为|(x/x+1)-2|>3，所以有x/x+1<-1，或x/x+1>5，所以有(2X+1)/X+1<0或(4X+5)/X+1<0
前一个不等式解集是-1<X<-0.5.后一个解集是
-5/4<X<-1，二者的并集即是原不等式的解集，为-5/4<X<-0.5且X不等于-1.

Answer 2

|(x/x+1)-2|>3
(x/x+1)-2>3 或 (x/(x+1)-2<-3
x/(x+1)>5 或 x/(x+1)<-1
x/x+1 -5>0 或 x/(x+1)+1<0
(x-5x-5)/(x+1)>0 或 (x+x+1)/(x+1)< 0
(-4x-5)/(x+1)>0 或 (2x+1)/(x+1)<0
(4x+5)/(x+1)<0 或 （2x+1)/(x+1)<0
-5/4<x<-1 或 -1<x<-1/2
所以 -5/4<x<-1/2 且 x≠-1

Answer 3

解：原式=
(x/x+1)-2>3或(x/(x+1)-2<-3
x/(x+1)>5 或x/(x+1)<-1
x/x+1 -5>0 或x/(x+1)+1<0
(x-5x-5)/(x+1)>0 或(x+x+1)/(x+1)< 0
(-4x-5)/(x+1)>0 或 (2x+1)/(x+1)<0
(4x+5)/(x+1)<0 或（2x+1)/(x+1)<0
-5/4<x<-1或-1<x<-1/2
所以-5/4<x<-1/2 且 x≠-1

Answer 4

|(x/x+1)-2|>3
x/(x+1)-2<-3,x/(x+1)-2>3

x/(x+1)-2<-3
x/(x+1)+1<0
(2x+1)/(x+1)<0
(2x+1)(x+1)<0
-1<x<-1/2

x/(x+1)-2>3
x/(x+1)-5>0
(x-5x-5)/(x+1)>0
(-4x-5)(x+1)>0
(4x+5)(x+1)<0
-5/4<x<-1

所以
-5/4<x<-1，-1<x<-1/2