Question

这个不等式怎么解啊？？

a(x-1)/(x-2)>1, (a>0)

希望可以详细一点，先谢谢了 :D

Answer 1

a(x-1)(x-2) > 1
(ax -a)/(x -2) -1 > 0
[(a-1)x + (2-a)]/(x -2) > 0

1. x > 2
(a-1)x + (2-a) > 0
(a - 1)x > a -2

A. a = 1
0 > a - 2 = 1-2 = -1恒成立
此时解为前提 x > 2

B. a > 1
x > (a -2)/(a -1)
因前提a > 1, x > 2, 很容易证明 (a-2)/(a-1) < 2
((a-2)/(a-1)为负数或小于1的正数)
此前提下解为: x > 2

C. a < 1
x < (a -2)/(a -1)
因前提a < 1, x > 2,
很容易证明 (a-2)/(a-1) -2 > 0; (a-2)/(a-1) > 2)
此前提下解为: 2 < x < (a -2)/(a -1)

2. x < 2
(a-1)x + (2-a) < 0
(a - 1)x < a -2

A. a = 1
0 < a - 2 = 1-2 = -1不成立
此时无解.

B. a > 1
x < (a -2)/(a -1)
因前提a > 1, x < 2, 很容易证明 (a-2)/(a-1) < 2
此前提下解为: x < (a -2)/(a -1)

C. a < 1
x > (a -2)/(a -1)
因前提a < 1, x < 2,
很容易证明 a-2)/(a-1) -2 > 0; (a-2)/(a-1) > 2
二者冲突，此时无解

Answer 2

x-2在分母，不等于0
所以(x-2)^2>0
两边乘以(x-2)^2
不等号不改向
a(x-1)(x-2)>(x-2)^2
(x-2)(ax-a-x+2)>0
(x-2)[(a-1)x-(a-2)]>0
a>2
a-1>0
所以(x-2)[x-(a-2)/(a-1)]>0

(a-2)/(a-1)=(a-1-1)/(a-1)=1-1/(a-1)
a>2,a-1>1
所以0<1/(a-1)<1
-1<-1/(a-1)<0
0<1-1/(a-1)<1
所以2>1-1/(a-1)
即2>(a-2)/(a-1)

所以x>2或x<(a-2)/(a-1)