Okay, so I need to find the real values of x that are not in the domain of the function f(x) = 1 divided by the sum of two absolute value expressions. The function is f(x) = 1 / (|x² + 3x - 4| + |x² + 9x + 20|). The domain of a function includes all the real numbers x for which the function is defined. Since this is a rational function with absolute values in the denominator, the only time the function would be undefined is if the denominator equals zero. So, the values of x that make the denominator zero are excluded from the domain. Therefore, my task is to solve the equation |x² + 3x - 4| + |x² + 9x + 20| = 0 and find all real solutions x. If there are such solutions, those x values are not in the domain. If the sum of the two absolute values is never zero, then the domain includes all real numbers.

First, I recall that the absolute value of any real number is always non-negative. Therefore, each term |x² + 3x - 4| and |x² + 9x + 20| is greater than or equal to zero. The sum of two non-negative numbers is zero only if both terms are zero simultaneously. So, the equation |x² + 3x - 4| + |x² + 9x + 20| = 0 implies that both |x² + 3x - 4| = 0 and |x² + 9x + 20| = 0 must hold true. Therefore, I can set each absolute value expression inside to zero and solve for x.

So, first, let's solve x² + 3x - 4 = 0. Then, solve x² + 9x + 20 = 0. The solutions to these equations will be the values of x that make each absolute value zero. If there exists an x that is a solution to both equations, then that x will make the denominator zero, and hence x is excluded from the domain. If there's no x that satisfies both equations, then the denominator is never zero, and the domain is all real numbers.

Let me start with the first quadratic equation: x² + 3x - 4 = 0. To solve this, I can use the quadratic formula. The quadratic formula is x = [-b ± √(b² - 4ac)] / (2a). Here, a = 1, b = 3, c = -4. Plugging these into the formula: discriminant D = 3² - 4*1*(-4) = 9 + 16 = 25. Since the discriminant is positive, there are two real solutions. The solutions are x = [-3 ± √25]/2 = [-3 ± 5]/2. Therefore, x = (-3 + 5)/2 = 2/2 = 1, and x = (-3 - 5)/2 = -8/2 = -4. So, the solutions are x = 1 and x = -4.

Next, let's solve the second quadratic equation: x² + 9x + 20 = 0. Again, using the quadratic formula. Here, a = 1, b = 9, c = 20. The discriminant D = 9² - 4*1*20 = 81 - 80 = 1. Since the discriminant is positive, there are two real solutions. The solutions are x = [-9 ± √1]/2 = [-9 ± 1]/2. Therefore, x = (-9 + 1)/2 = -8/2 = -4, and x = (-9 - 1)/2 = -10/2 = -5. So, the solutions are x = -4 and x = -5.

Now, we need to check if there's any x that is a solution to both equations. From the first equation, the solutions are x = 1 and x = -4. From the second equation, the solutions are x = -4 and x = -5. The only common solution is x = -4. Therefore, x = -4 is the value that makes both absolute value expressions zero. Therefore, when x = -4, the denominator becomes 0 + 0 = 0, which makes f(x) undefined. Therefore, x = -4 is not in the domain of f(x). The other solutions, x = 1 and x = -5, make only one of the absolute value expressions zero. For example, x = 1 makes the first absolute value zero but the second absolute value |1² + 9*1 + 20| = |1 + 9 + 20| = |30| = 30, which is positive. Similarly, x = -5 makes the second absolute value zero but the first absolute value |(-5)^2 + 3*(-5) -4| = |25 -15 -4| = |6| = 6, which is positive. Therefore, only x = -4 causes the denominator to be zero.

But wait, let me verify this. Let's check x = -4 in the denominator: |x² + 3x -4| + |x² + 9x +20|. Plugging in x = -4: |(-4)^2 + 3*(-4) -4| + |(-4)^2 +9*(-4) +20| = |16 -12 -4| + |16 -36 +20| = |0| + |0| = 0 + 0 = 0. So, yes, denominator is zero. For x =1: |1 +3 -4| + |1 +9 +20| = |0| + |30| = 0 +30=30. For x=-5: |25 -15 -4| + |25 -45 +20| = |6| + |0| =6+0=6. So, both are non-zero. Therefore, only x=-4 makes the denominator zero. Therefore, the only value excluded from the domain is x=-4.

Therefore, the answer should be x = -4. So, the real value not in the domain is -4. Therefore, the final answer is \boxed{-4}.

But wait, let me just double-check if there's any other way the denominator could be zero. For example, could the two absolute values somehow cancel each other? But since absolute values are non-negative, the sum of two non-negative numbers can only be zero if both are zero. Therefore, unless both expressions inside the absolute values are zero, the sum is positive. Therefore, the only solution is when both quadratics are zero, which only happens at x=-4. Therefore, that's the only excluded value.

Therefore, the conclusion is correct. The domain excludes x = -4. Therefore, the answer is \boxed{-4}.
So, after careful consideration and solving the equations step by step, the only real value of \( x \) that is not in the domain of the function \( f(x) \) is \( x = -4 \). Therefore, the final answer is \boxed{-4}.
</think>