Okay, let's tackle this inequality: |x² - 81| divided by (x² - 36x) is less than zero. Hmm, so we need to find all x values where this fraction is negative. Let me think step by step.

First, I remember that an inequality involving a fraction being less than zero means the numerator and denominator must have opposite signs. So either the numerator is positive and the denominator is negative, or the numerator is negative and the denominator is positive. Since the numerator is an absolute value, which is always non-negative, that simplifies things a bit. Wait, the absolute value is always non-negative, so |x² - 81| is either zero or positive. But since the entire fraction must be less than zero, the numerator can't be zero because that would make the fraction zero, which isn't less than zero. Therefore, the numerator must be positive, and the denominator must be negative. So the key is to find where the denominator is negative.

Wait, let me confirm that. If the numerator is positive (since absolute value is non-negative) and the denominator is negative, then the fraction is negative. If the numerator is zero (which would make the fraction zero), that's not less than zero. So yes, we need the numerator to be positive and the denominator to be negative. Therefore, the denominator must be negative.

So first, let's factor the denominator: x² - 36x. Factor out an x: x(x - 36). So the denominator is x(x - 36). We need to find where this is negative. The critical points are at x = 0 and x = 36. So we can test intervals around these points.

The denominator x(x - 36) is a quadratic opening upwards. It will be negative between its roots. So between 0 and 36, the denominator is negative. So x in (0, 36) makes denominator negative.

Now, the numerator is |x² - 81|. Let's analyze when this is positive. The absolute value is positive except when x² - 81 = 0, which is when x = ±9. So numerator is positive when x ≠ ±9. Therefore, the fraction is positive when denominator is positive and x ≠ ±9, and negative when denominator is negative and x ≠ ±9. But since we need the fraction to be less than zero, we take the case where denominator is negative, which is x in (0, 36), but excluding points where numerator is zero, which is x = ±9. Wait, but x = 9 is in (0, 36), so x = 9 would make numerator zero, making the fraction zero. Similarly, x = -9 is not in (0, 36), so that's okay.

Therefore, the solution set is x in (0, 36) excluding x = 9. But wait, let me make sure. Let's check the critical points.

First, denominator is zero at x = 0 and x = 36. These are excluded because the original expression is undefined there. The numerator is zero at x = ±9. At x = 9, the numerator is zero, so the fraction is zero, which doesn't satisfy the inequality. So we have to exclude x = 9 from the interval (0, 36). So the solution is (0, 9) union (9, 36).

But wait, let's verify. Let's pick a test point in (0, 9), say x = 1. Denominator is 1*(1 - 36) = -35, which is negative. Numerator is |1 - 81| = 80, which is positive. So 80 / (-35) is negative, which satisfies the inequality. Good.

Another test point in (9, 36), say x = 18. Denominator is 18*(18 - 36) = 18*(-18) = -324, negative. Numerator is |324 - 81| = 243, positive. 243 / (-324) is negative. Also satisfies.

At x = 9, numerator is zero, so fraction is zero, which is not less than zero. So exclude 9.

What about x approaching 0 from the right? Let's say x = 0.1. Denominator is 0.1*(0.1 - 36) = 0.1*(-35.9) = -3.59, negative. Numerator is |0.01 - 81| = 80.99, positive. So negative. So x approaching 0 from the right is okay.

What about x approaching 36 from the left? Let's say x = 35.9. Denominator is 35.9*(35.9 - 36) = 35.9*(-0.1) = -3.59, negative. Numerator is |(35.9)^2 - 81|. Let's compute (35.9)^2: 35^2 = 1225, 0.9^2 = 0.81, cross term 2*35*0.9 = 63. So (35 + 0.9)^2 = 35^2 + 2*35*0.9 + 0.9^2 = 1225 + 63 + 0.81 = 1288.81. So |1288.81 - 81| = 1207.81, positive. So fraction is negative. So x approaching 36 from the left is okay.

So the interval (0, 36) excluding 9 is the solution. But wait, let's check x = 36. At x = 36, denominator is zero, so undefined. So not included.

But hold on, the original inequality is |x² - 81| / (x² - 36x) < 0. Let's also check if there are any other points where denominator is negative. The denominator is x(x - 36), so negative between 0 and 36. So that's the only interval where denominator is negative. So the numerator is always positive except at x = ±9, which are within that interval. So the only exclusion is x = 9.

Therefore, the solution is x ∈ (0, 9) ∪ (9, 36). So in interval notation, that's two intervals: from 0 to 9 and from 9 to 36.

But let me check if there's any other point where the numerator is zero. The numerator |x² - 81| is zero when x² - 81 = 0, so x = ±9. So we already considered that. So the only excluded point in the denominator's negative interval is x = 9.

Wait, but x = -9 is not in the interval (0, 36), so it's okay. So the final answer is (0, 9) union (9, 36).

But let me also check if there's any other way this could be approached. For example, sometimes when dealing with absolute values, you can split into cases. Let's try that to confirm.

Case 1: x² - 81 ≥ 0. Then |x² - 81| = x² - 81. So the inequality becomes (x² - 81)/(x² - 36x) < 0. So we need (x² - 81)(x² - 36x) < 0. Wait, but since denominator is x² - 36x, we have (x² - 81)/(x(x - 36)) < 0.

Alternatively, perhaps it's better to analyze when (x² - 81)/(x(x - 36)) is negative. But since the numerator is |x² -81|, which is x² -81 when x² -81 ≥0, i.e., x ≥9 or x ≤-9. But in the denominator's interval (0, 36), x is positive, so x ≥9 is part of that interval. So let's see.

Case 1: x² -81 ≥0 → x ≥9 or x ≤-9. But in the interval (0, 36), x can't be ≤-9, so only x ≥9. Then in this case, numerator is x² -81, denominator is x(x -36). So the expression becomes (x² -81)/(x(x -36)) <0. Since in this case x ≥9, so x is positive, and x -36 is negative (since x <36 in the denominator's interval). Therefore, denominator is positive times negative, so negative. Numerator x² -81: when x ≥9, x² -81 ≥0. So positive numerator, negative denominator: positive/negative = negative. So the inequality holds for all x ≥9 in the denominator's interval. But wait, the denominator's interval is (0,36). So x in (9,36). So in this case, the inequality holds here.

Case 2: x² -81 <0 → |x² -81| = -(x² -81) = 81 -x². So the inequality becomes (81 -x²)/(x² -36x) <0. Let's factor numerator: 81 -x² = -(x² -81) = -(x -9)(x +9). Denominator is x(x -36). So the expression becomes [-(x -9)(x +9)]/[x(x -36)] <0. Let's simplify this: multiply numerator and denominator by -1 (remembering to flip the inequality sign, but since we're dealing with <0, the direction remains the same). So [ (x -9)(x +9) ] / [x(x -36)] >0. Now, we need to find where this expression is positive.

But let's analyze the intervals. The critical points are at x = -9, 0, 9, 36. Since the original denominator is x(x -36), and in this case, we are considering x² -81 <0, which is x ∈ (-9,9). But in the denominator's interval (0,36), we are considering x ∈ (0,36). So in this case, x ∈ (0,9). So let's see.

In the numerator, (x -9)(x +9): when x is in (0,9), x +9 is positive (since x >0), x -9 is negative. So numerator is positive * negative = negative. Denominator: x is positive, x -36 is negative. So denominator is positive * negative = negative. Therefore, the expression is [negative]/[negative] = positive. So the inequality [ (x -9)(x +9) ] / [x(x -36)] >0 is true for x in (0,9). Therefore, since in Case 2, we have x in (0,9), the inequality holds here. So combining both cases:

From Case 1: x in (9,36)

From Case 2: x in (0,9)

Excluding x =9 where the expression is zero. Therefore, the solution is (0,9) ∪ (9,36).

So this confirms the earlier result. Therefore, the final answer is x ∈ (0,9) ∪ (9,36). So in boxed notation, we can write it as (0, 9) ∪ (9, 36). Let me check if there's any other way to write this. Alternatively, since it's a union of two open intervals, that's the standard way.

Wait, but let me check if there's a point where the numerator is zero. For example, x =9: numerator is |81 -81| =0, so the fraction is zero, which is not less than zero. So x=9 is excluded. Similarly, x=-9: but x=-9 is not in the interval (0,36), so it's okay. So the final answer is indeed (0,9) union (9,36). So in LaTeX, that would be \boxed{(0, 9) \cup (9, 36)}.

But just to be thorough, let's check another point in (0,9), say x=5. Numerator: |25 -81|=56. Denominator: 5*(5-36)=5*(-31)=-155. So 56 / (-155) is negative. Correct.

Another point in (9,36), say x=20. Numerator: |400 -81|=319. Denominator: 20*(20-36)=20*(-16)=-320. 319/-320 is negative. Correct.

At x=18, same as before, negative. At x approaching 0 from the right, denominator approaches 0 from negative, numerator approaches 81, so negative. Correct.

At x approaching 36 from the left, denominator approaches 0 from negative, numerator approaches |36² -81|=|1296 -81|=1215, so fraction approaches negative infinity. Correct.

Therefore, all checks pass. So the solution is correct.
Therefore, the solution to the inequality is the union of the intervals $(0, 9)$ and $(9, 36)$.

\[\boxed{(0, 9) \cup (9, 36)}\]
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