Solve: x-(3/x)<0

is the answer(-infinity, -sqrt3)U(0,sqrt3)
?

x<+-sqrt3

Now if it is lees than +sqrt3, it is also less than -sqrt3.

Did I find this out correctly?

To solve the inequality x - (3/x) < 0, we can follow these steps:

Step 1: Multiply through by x to eliminate the fraction:
x^2 - 3 < 0

Step 2: Bring all terms to one side:
x^2 - 3 - 0 < 0
x^2 - 3 < 0

Step 3: Factor the quadratic expression, if possible:
(x + sqrt(3))(x - sqrt(3)) < 0

Step 4: Determine the sign of each factor:
For (x + sqrt(3)):
Since sqrt(3) is a positive number, (x + sqrt(3)) is positive when x > -sqrt(3) and negative when x < -sqrt(3).
For (x - sqrt(3)):
Similarly, (x - sqrt(3)) is positive when x > sqrt(3) and negative when x < sqrt(3).

Step 5: Analyze the intervals:
We need to look at the intervals where the inequality is true, i.e., when the expression (x + sqrt(3))(x - sqrt(3)) is negative.

For (x + sqrt(3))(x - sqrt(3)) < 0 to be true, we have two cases:
- When (x + sqrt(3)) < 0 and (x - sqrt(3)) > 0, which means -sqrt(3) < x < sqrt(3).
- When (x + sqrt(3)) > 0 and (x - sqrt(3)) < 0, but since (x + sqrt(3)) is always positive, this case is not possible.

Step 6: Combine the intervals:
Therefore, the solution to the inequality x - (3/x) < 0 is (-sqrt(3), sqrt(3)).

In summary, the answer to the inequality is (-sqrt(3), sqrt(3)), or in interval notation: (-∞, -sqrt(3)) ∪ (0, sqrt(3)).