<Question>
Solve
\[\frac{|x^2 - 81|}{x^2 - 36x} < 0.\]
</Question>

<think>
<Plan_1>
Okay, so I need to solve the inequality |x² - 81| divided by (x² - 36x) is less than zero. Let me write that down again to make sure I have it right: |x² - 81| / (x² - 36x) < 0. Alright, so the numerator is an absolute value, which is always non-negative, right? And the denominator is a quadratic expression. Hmm, so for the entire fraction to be less than zero, the numerator has to be positive, and the denominator negative. Because absolute value is non-negative, so if the denominator is negative, the whole fraction will be negative. But if the numerator is zero, then the fraction is zero, which doesn't satisfy the inequality. So, first, let me note that the numerator |x² - 81| is zero when x² - 81 = 0, which is when x = 9 or x = -9. So at x = 9 and x = -9, the expression is zero, which is not less than zero, so those points are excluded from the solution.

Denominator: x² - 36x.
Let me factor that: x(x - 36). So the denominator is zero when x = 0 or x = 36. Therefore, the expression is undefined at x = 0 and x = 36, so those points are also excluded.

So, the domain of the expression is all real numbers except x = -9, 0, 9, 36. Wait, no: the numerator is |x² -81|, which is defined for all real numbers, but the denominator can't be zero, so x ≠ 0 and x ≠36. Also, the numerator is zero at x = -9 and x =9, so those points are excluded as well because the expression would be zero there.
So the domain is ℝ \ { -9, 0, 9, 36 }.

Now, the inequality is |x² -81| / (x(x -36)) < 0.

Since the numerator is an absolute value, it's always non-negative. So the sign of the entire expression depends on the denominator. But wait, but the numerator is |x² -81|, which is zero only when x² -81 is zero, but we already excluded those points. So, except at x = -9 and x =9, the numerator is positive. Therefore, the fraction is positive when denominator is positive, and negative when denominator is negative. Therefore, the inequality |x² -81| / (x² -36x) <0 is equivalent to denominator <0.

Therefore, solving x² -36x <0.

<prompt_1.1> Simplify the inequality to focus on the denominator. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's solve that inequality. x² -36x <0. Factor out x: x(x -36) <0. So, critical points at x=0 and x=36. Testing intervals:

1. x <0: pick x=-1. (-1)(-1 -36)= (-1)(-37)=37 >0. So positive.

2. 0 <x <36: pick x=1. 1(1 -36)=1*(-35)=-35 <0. Negative.

3. x >36: pick x=37. 37(37 -36)=37*1=37>0. Positive.

So the inequality x(x -36) <0 is true when 0 <x <36. So the denominator is negative when x is between 0 and 36. Therefore, the original inequality is equivalent to 0 <x <36, but we have to remember the domain restrictions. The original expression is undefined at x=0 and x=36, so those endpoints are excluded. Also, we need to check if there are any other restrictions.

But wait, the numerator is |x² -81|, which is positive except at x= -9 and x=9, but in the interval 0 <x <36, x can't be -9, so the numerator is positive except at x=9. But at x=9, the numerator is zero, so the entire expression is zero, which doesn't satisfy the inequality. Therefore, in the interval (0,36), excluding x=9.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but let me check: if x is in (0,36), then denominator is negative, numerator is |x² -81|, which is positive except when x²=81, i.e., x=9 or x=-9. But in (0,36), x=9 is included, but at x=9, numerator is zero, so expression is zero, which is not less than zero. Therefore, we need to exclude x=9 as well.

Therefore, the solution would be (0,36) excluding 9. So (0,9) union (9,36).

But wait, but let me make sure there are no other points where the numerator could be zero in that interval. The numerator |x² -81| is zero when x²=81, so x=9 or x=-9. So only x=9 is in (0,36), so only x=9 is excluded. So the solution is (0,9) union (9,36).

But let me check another way. Let's consider the original inequality: |x² -81| / (x² -36x) <0.

<prompt_2.1> Verify the solution by considering test points in different intervals. </prompt_2.1>

But wait, let me check the numerator again. The numerator is |x² -81|, which is non-negative, but when is it positive? When x² ≠81, so x≠9 and x≠-9.

<prompt_2.2> Re-evaluate the solution by considering the positivity of the numerator. </prompt_2.2>

But let me check another test point in (0,9), say x=1.

<prompt_2.3> Confirm the solution by testing additional points. </prompt_2.3>

But wait, but let me check x approaching 0 from the right.

<prompt_2.4> Analyze the behavior of the expression near boundary points. </prompt_2.4>

Similarly, as x approaches 36 from the left, denominator approaches 36*(36 -36)=0, but from the left, denominator x(x-36) approaches 36*(-0) which is 0 from negative side, so the expression approaches negative infinity.

<prompt_2.5> Verify the behavior of the expression near boundary points. </prompt_2.5>
</Plan_2>

<Execution_2>
<execution_2.1>
The numerator is |x² -81|, which is non-negative, so the sign of the entire expression is determined by the denominator. So when denominator is negative, the expression is negative, which satisfies the inequality. So when x is in (0,36), denominator is negative, so the inequality holds. However, at x=9, numerator is zero, so the expression is zero, which does not satisfy the strict inequality. Therefore, the solution is (0,36) excluding 9, which is (0,9) union (9,36). But let me check if there's any other point where the numerator is zero in this interval. As mentioned, only x=9. So that's the only exclusion.
</execution_2.1>

<execution_2.2>
So |x² -81| = x² -81 when x² >=81, and 81 -x² when x² <81. So, depending on x, the numerator can be written differently. Maybe I need to split into cases based on x² -81 being positive or negative.

Case 1: x² -81 >=0, which is x <= -9 or x >=9.

Case 2: x² -81 <0, which is -9 <x <9.

So maybe I need to consider these cases separately.

So first, let's split into cases based on the numerator.

Case 1: x <= -9 or x >=9. Then |x² -81| = x² -81. So the inequality becomes (x² -81)/(x² -36x) <0.

Case 2: -9 <x <9. Then |x² -81| =81 -x². So inequality becomes (81 -x²)/(x² -36x) <0.

So perhaps splitting into these cases would be more accurate.
</execution_2.2>

<execution_2.3>
Then |1 -81| = | -80 | =80, positive. Denominator 1 -36= -35. So 80/-35 ≈-2.2857 <0. So satisfies the inequality.

At x=10, which is in (9,36). Numerator |100 -81|=19, positive. Denominator 10(10 -36)=10*(-26)= -260. So 19/-260≈-0.073 <0. So satisfies the inequality. But x=9, numerator is 0, so expression is 0, which doesn't satisfy.

What about x=45? Then numerator |2025 -81|=1944, denominator 45(9)=405, so 1944/405≈4.8 >0. So doesn't satisfy. So outside (0,36), denominator is positive or negative? Wait, x=45 is greater than 36, denominator is positive. So between 36 and infinity, denominator is positive, so expression is positive. So the inequality is satisfied only between 0 and 36, excluding 9. But let me check x= -1, which is less than 0. Then numerator |1 -81|=80, denominator (-1)(-37)=37. So 80/37≈2.16>0. So doesn't satisfy.

So yes, seems like only between 0 and 36, excluding 9.
</execution_2.3>

<execution_2.4>
At x=1, numerator |1 -81|=80, denominator 1 -36= -35, so 80/-35 <0. So true. At x=2, same thing, numerator |4 -81|=77, denominator 2 -72= -70, 77/-70 <0. So true. At x=8, numerator |64 -81|=17, denominator 8 - 288= -280, 17/-280 <0. True. So in (0,9), it's true. At x=10, which is in (9,36), numerator |100 -81|=19, denominator 10 - 360= -350, 19/-350 <0. So also true. Wait, but according to previous reasoning, x=9 is excluded. So in (0,36) excluding x=9.
</execution_2.4>

<execution_2.5>
But since we already have the denominator negative in (0,36), and numerator positive, so the expression is negative there, so the inequality holds. So yes, (0,36)\{9}.

But wait, but let me check another point. Let's take x=10, which is in (9,36). Then denominator is 10*(10 -36)=10*(-26)=-260 <0. Numerator |10² -81|=|100 -81|=19 >0. So the expression is 19/-260 <0, which satisfies the inequality. So x=10 is included. But at x=9, numerator is zero, so expression is zero, which is not less than zero. So x=9 must be excluded.

Similarly, at x=0, denominator is zero, so excluded. At x=36, denominator zero, excluded. So the solution is (0,36) excluding 9, so (0,9) union (9,36).
</execution_2.5>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2, execution_2.3, execution_2.4, execution_2.5:
But let me verify with another approach. Let's consider the original inequality:

<prompt_3.1> Confirm the solution using the alternative approach. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
|x² -81| / (x² -36x) <0.

Since the numerator is absolute value, which is non-negative, and the denominator is x² -36x. So the fraction is negative when denominator is negative (since numerator is positive except at zeros). Therefore, as before, denominator <0. So x² -36x <0. Which is x(x -36) <0. Solution is 0 <x <36. But we have to exclude x=0 and x=36 where denominator is zero. Also, check if numerator is zero in this interval. The numerator is zero when x²=81, so x=9 or x=-9. In 0 <x <36, only x=9 is present. So at x=9, the expression is zero, which doesn't satisfy the inequality. Therefore, solution is (0,36)\{9} = (0,9) union (9,36). So same as before.

Therefore, seems like the solution is (0,9) union (9,36).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me check x= -15, which is in Case 1 (x <= -9). Then |x² -81| = x² -81. So numerator is (-15)^2 -81=225 -81=144.

<prompt_4.1> Finalize the solution by checking additional points and confirming the result. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Denominator: (-15)(-15 -36)= (-15)(-51)=765. So 144/765 >0. So doesn't satisfy the inequality. So in x <= -9, the expression is positive, so not satisfying the inequality. So only in -9 <x <9, but wait, in -9 <x <9, the numerator is 81 -x², denominator x(x -36). Let's check x= -5.

Case 2: -9 <x <9. Let's take x=-5. Then numerator |25 -81|=56. Denominator (-5)(-5 -36)= (-5)(-41)=205. So 56/205 >0. Doesn't satisfy. x=5. Numerator |25 -81|=56. Denominator 5(5 -36)=5*(-31)= -155. So 56/-155 <0. So satisfies. So in 0 <x <9, denominator is negative, numerator positive, so expression negative. So inequality holds. So seems like (0,9) union (9,36). But why in Case 2, when -9 <x <0, the denominator is positive (since x is negative and x -36 is negative, so product positive), and numerator is positive, so expression positive. So inequality not satisfied. So only in 0 <x <36, denominator negative, numerator positive, so expression negative. But at x=9, numerator zero. So yes, the solution is (0,9) union (9,36).

Therefore, after thorough checking, the solution is x in (0,9) union (9,36).
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
**Final Answer**
The solution to the inequality is \boxed{(0, 9) \cup (9, 36)}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
\[x \in \boxed{(0, 9) \cup (9, 36)}.
\]
</Final_answer>
</think>