Question

解关于x的不等式. [a(x-1)]/(x-2)的解集是A,且3不属于A.求实数A的取直范围/求集合A

Answer 1

应该是

[a(x-1)]/(x-2)>2

吧

一、化简：

[a(x-1)]/(x-2)>2

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

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

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

(ax-a-2x+4)/(x-2)>0

[(a-2)x-(a-4)]/(x-2)>0

二、讨论a的值

因为[a(x-1)]/(x-2)>2的结集为A,且3不属于A

所以当x=3时

[(a-2)x-(a-4)]/(x-2)>0

[3(a-2)-(a-4)]/(3-2)>0

3(a-2)-(a-4)>0

3a-6-a+4>0

2a-2>0

a>1

因为x≠3(补集的思想)

所以a≤1

三、讨论x值

因为[(a-2)x-(a-4)]/(x-2)>0

所以a-2≤-1