Question

解不等式x+ (1/2) <1/(x-1) -1≤x+(1/2)≤1 -1≤1/(x-1)≤1 三者联立 问x范围

能有详细解答过程吗

Answer 1

解：
-1≤x+(1/2)≤1
-1-1/2≤x≤1-1/2
-3/2≤x≤1/2

-1≤1/(x-1)≤1
由1/(x-1)≤1得x≥2或x<1
由-1≤1/(x-1)得x>1或x≤0
∴x≥2或x≤0
以上两式的交集为-3/2≤x≤0
∴x-1<0

x+ (1/2) <1/(x-1)
2x+1<2/(x-1)
(2x+1)(x-1)>2
2x²-x-1>2
2x²-x-3>0
(2x-3)(x+1)>0
x>3/2或x<-1
与-3/2≤x≤0取交集得-3/2≤x<-1
综上，所求x的范围为-3/2≤x<-1