Question

-1/2＜(m+1)/(2m-1)≤1中m的取值范围?

Answer 1

-1/2＜(m+1)/(2m-1)≤1

-1/2＜(m+1)/(2m-1)
[2m-1+2(m+1)]/(2m-1)＞0
(4m+1)/(2m-1)＞0
(4m+1)(2m-1)＞0
m＜-1/4或m＞1/2

(m+1)/(2m-1)≤1
(2m-1-m-1)/(2m-1)≥0
(m-2)/(2m-1)≥0
(m-2)(2m-1)≥0,m≠1/2
所以m＜1/2或m≥2

取'm＜-1/4或m＞1/2'与'm＜1/2或m≥2'的交集得
m＜-1/4或m≥2

即解集是{m|m＜-1/4或m≥2}

Answer 2

(m+1)/(2m-1)
=(m-1/2+3/2)/(2m-1)
=1/2+(3/2)/(2m-1)

-1/2<1/2+(3/2)/(2m-1)≤1
-1<(3/2)/(2m-1)≤1/2
-2/3<1/(2m-1)≤1/3
-2/3<1/(2m-1)<0或0<1/(2m-1)≤1/3
2m-1<-3/2或2m-1≥3
∴ 2m<-1/2或2m≥4
∴ m< -1/4或 m≥2

Answer 3

-1/2＜(m+1)/(2m-1)≤1
(m+1)/(2m-1)≤1
(m+1-2m+1)/(2m-1)≤0
(2-m)(2m-1)≤0
(m-2)(2m-1)≥0
得 m≥2 或 m<1/2

(m+1)/(2m-1)>-1/2
(m+1+m-1/2)/(2m-1)>0
(2m+1/2)(2m-1)>0
得 m>1/2 或 m<-1/4
取交集得
m≥2 或 m<-1/4