Question

帮解个分式不等式

(X-5)处以（X-1）小于等于2

把过程详细的写出来

Answer 1

由题意,x不等于1
若x>1,则x-1>0
不等式两边同时乘以x-1，符号不变
x-5<=2(x-1)
x-5<=2x-2
合并同类项
x>=-3
再根据假设有x>1
若x<1
不等式两边同时乘以x-1，符号改变
x-5>=2(x-1)
x-5>=2x-2
合并同类项
x<=-3

综上所述
x<=-3或x>1

Answer 2

由题意,x不等于1
若x>1,则x-1>0,(x-5)/(x-1)<=2
不等式两边同时乘以x-1，符号不变
(x-5)<=2(x-1)
x-5<=2x-2
-5+2<=2x-x
-3<=x
即x>=-3
若x<1
不等式两边同时乘以x-1，符号改变
(x-5)>=2(x-1)
x-5>=2x-2
-5+2<=2x-x
-3>=x
即x<=-3

所以x<=-3或x>1

Answer 3

(x-5)/(x-1)≤2
移项,得
(x-5)/(x-1)-2≤0
[(x-5)-2(x-1)]/(x-1)≤0
-(x+3)/(x-1)≤0
(x+3)/(x-1)≥0
用数轴标根法可解得
x≤-3或x≥1
又x-1为分母
∴x≠1
∴x≤-3或x>1

Answer 4

解:
(x-5)/(x-1)≤2
(x-5)/(x-1)-2≤0
(x-5)/(x-1)-2(x-1)/(x-1)≤0
(x+3)/(x-1)≥0
其等价于
(x+3)(x-1)≥0,且x-1≠0
所以，x>1或x≤-3