solution: $x$ is not in the domain of $f$ if the denominator is zero.  Since both absolute values are nonnegative, both must be zero for the denominator to be zero.  So
  
  \begin{align*}
  0=x^2+3x-4=(x+4)(x-1)&\Rightarrow x=-4\text{ or }x=1\\
  0=x^2+9x+20=(x+4)(x+5)&\Rightarrow x=-4\text{ or }x=-5
  \end{align*}
  
  The only value of $x$ which makes both absolute values zero is $x=\boxed{-4}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  What real values of $x$ are not in the domain of
  
  $f(x)=\frac{1}{|x^2+3x-4|+|x^2+9x+20|}$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: -4
extra_info.question: What real values of $x$ are not in the domain of
  
  $f(x)=\frac{1}{|x^2+3x-4|+|x^2+9x+20|}$?
extra_info.split: test