Question

解不等式1<|2x+1|≤3,并求出其解集.

Answer 1

　不等式1<|2x+1|≤3
(1)2X+1>0,X>-1/2
1<2X+1<=3
0<2X<=2
0<X<=1
即有0<X<=1
(2)2X+1<0,X<-1/2
1<-(2X+1)<=3
-3<=2x+1<-1
-4<=2x<-2
-2<=x<-1
即有-2<=x<-1
综上有解集是{x|-2<=x<-1,0<x<=1}

Answer 2

分两步:
(1)|2x+1|≤3，∴-3≤2x+1≤3,∴-2≤x≤1
(2)1<|2x+1│，∴2x+1＞1或2x+1＜-1.∴x＞0或x＜-1
综上可知：-2≤x＜-1或0＜x≤1
∴其解集是﹛x│-2≤x＜-1或0＜x≤1﹜

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Answer 3

解析

2x+1>1 (1)
2x+1<-1 (2）

x>0或
x<-1

2x+1≤3
或2x+1≥-3
解 x≤1
x≥-2
所以大大取大，小小取小，
【-2 -1）∪（0 1】

希望对你有帮助
学习进步O(∩_∩)O谢谢

Answer 4

由|2x+1|>1，则2x+1>1或2x+1<-1，分别求得x>0或x<-1；由|2x+1|≤3，得-3≤2x+1≤3，解得-1≤x≤1。综合可得，解为0<x≤1

Answer 5

1<|2x+1|≤3
-3≤2x+1<-1或1<2x+1≤3
-2≤x<-1或0<x≦1
∴解集是x∈[-2,-1)∪(0,1]