<Question>
What real values of $x$ are not in the domain of

$f(x)=\frac{1}{|x^2+3x-4|+|x^2+9x+20|}$?
</Question>

<think>
<Plan_1>
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by (the absolute value of x² + 3x - 4 plus the absolute value of x² + 9x + 20). Hmm, the domain of a function includes all the real numbers x for which the function is defined. Since this is a reciprocal function, the main thing to watch out for is division by zero. That means I need to find where the denominator is zero because dividing by zero is undefined. So, my task is to find the values of x that make the denominator zero and exclude them from the domain.
Let me write down the denominator to focus: |x² + 3x - 4| + |x² + 9x + 20|. The sum of two absolute values. Since absolute values are always non-negative, their sum is also non-negative. However, the sum could be zero only if both absolute values are zero simultaneously. Because if either one is positive, the other being non-negative would make the sum positive. Therefore, the denominator is zero only when both |x² + 3x - 4| = 0 and |x² + 9x + 20| = 0. Which would require that x² + 3x - 4 = 0 and x² + 9x + 20 = 0. So, I need to solve these two equations simultaneously.

First, let's solve x² + 3x - 4 = 0.

<prompt_1.1> Solve the equation x² + 3x - 4 = 0. </prompt_1.1>

Now, solving x² + 9x + 20 = 0.

<prompt_1.2> Solve the equation x² + 9x + 20 = 0. </prompt_1.2>
-----

But wait, let me check if there's another approach. Suppose someone doesn't factor the quadratics but instead tries to analyze the denominator. Let's see.

<prompt_1.3> Explore an alternative method to confirm the solution. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me factor that. Looking for two numbers that multiply to -4 and add to 3. Hmm, 4 and -1? Yes, (x + 4)(x - 1) = x² + 3x - 4. So, the solutions are x = -4 and x = 1.
</execution_1.1>

<execution_1.2>
Let me factor that. 20 factors into 4 and 5, and 4 + 5 is 9. So, (x + 4)(x + 5) = x² + 9x + 20. Therefore, the solutions are x = -4 and x = -5.
</execution_1.2>

<execution_1.3>
The denominator is |x² + 3x -4| + |x² +9x +20|. Since absolute values are non-negative, their sum is zero only if each absolute value is zero. So, indeed, we need both x² +3x -4 =0 and x² +9x +20=0. Therefore, the previous conclusion holds.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
So, the solutions to the first equation are x = 1 and x = -4, and the solutions to the second equation are x = -4 and x = -5. The common solution here is x = -4. Therefore, x = -4 is the only value where both quadratics are zero. For all other x, at least one of the quadratics is non-zero, so their absolute values would add up to a positive number, making the denominator positive, which is allowed. Hence, the only value excluded from the domain is x = -4.

Wait a minute, let me verify this. Suppose there is another x where one of the quadratics is positive and the other is negative, but their absolute values might cancel each other out?

<prompt_2.1> Verify the solution by considering if the sum of absolute values can be zero for other x values. </prompt_2.1>
-----

Based on execution_1.3:
Therefore, the answer is x = -4. 

Just to recap: The denominator is the sum of two absolute values. For the sum to be zero, both absolute values must be zero. Each absolute value expression equals zero only at specific x-values. The intersection of those x-values is the solution.

<prompt_2.2> Recap the solution and confirm the final answer. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
But since they are absolute values, they can't be negative, so their sum can't be zero unless both are zero. Therefore, even if one is positive and the other is negative, their absolute values would add up to a positive number. Therefore, the only x that makes both quadratics zero is x = -4.
</execution_2.1>

<execution_2.2>
In this case, only x = -4 is common to both, so that's the only x excluded from the domain. 

I think that's thorough.
</execution_2.2>
</Execution_2>

<Final_answer>
Therefore, the real value of $x$ that is not in the domain of $f(x)$ is $\boxed{-4}$.
</Final_answer>
</think>