When does mathematics solve inequality take intersection and union?

Generally speaking, there is an OR relationship between them, and an AND relationship between them, which is an intersection. That is, if two or more of them are various situations of an event, the results are merged; If one of the two is the premise of the other, it is merged.

However, this problem is not a simple combination.

The first | (x+3)/(x-1) | >; 1

Then (x+3)/(x- 1)> 1 or (x+3)/(x- 1)

These are two possible situations. I don't need to tell you the solution. Just combine the results of the two inequalities (note: the sign of x- 1, that is, its sign, must be considered when eliminating the denominator).

The second |x-3| >x- 1

You can't directly use the formula | x | & gta(a & gt；; 0) then x>a or x < -a

If you want to use it directly, then x- 1 must be less than 0. Because for any x, |x| >=0 exists. If the right side of the inequality sign is less than 0, it need not be considered. Of course, you can't use it directly, otherwise there may be a lack of solutions.

But at this time, the value of x is unknown, so it cannot be said that x- 1 must be greater than 0. How should this problem be solved?

Take the method of case discussion:

Case1:x-1> 0

Then x-3 >; X- 1 or x-3

Because the prerequisite is X- 1 >: 0.

Therefore, the solution of the case 1 is 1

Case 2: x- 1=0

Then x= 1

|x-3|=|-2| >0=x- 1

At this time, X= 1.

Case 3: X- 1

Then |x-3| >X- 1 hold. At this time, X.