solution: The function is defined when the value inside the square root is positive, i.e. we must have $x^2-5x+6>0$.  Factoring, we get $(x-3)(x-2)>0$. So either both factors in the left hand side are negative or they are both positive. They are both negative when $x<2$. They are both positive when $x>3$. So the domain of $f(x)$ is $x<2 \text{ or } x>3$, or $x \in \boxed{(-\infty, 2) \cup (3, \infty)}$ in interval notation.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  What is the domain of the real-valued function $f(x)=\frac{2x-7}{\sqrt{x^2-5x+6}}$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: (-\infty, 2) \cup (3, \infty)
extra_info.question: What is the domain of the real-valued function $f(x)=\frac{2x-7}{\sqrt{x^2-5x+6}}$?
extra_info.split: test