Okay, so I need to find the domain of the real-valued function f(x) = (2x - 7)/sqrt(x² - 5x + 6). Hmm, let me recall how to determine the domain of a function. The domain is all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive (because you can't take the square root of a negative number and it would be a real number), and also that the denominator isn't zero because division by zero is undefined. 

So, first, let's look at the denominator: sqrt(x² - 5x + 6). For the square root to be defined and non-zero, the radicand (the expression inside the square root) must be greater than zero. That is, x² - 5x + 6 > 0. 

Wait, but the numerator is 2x - 7, which is a linear expression. However, the numerator doesn't affect the domain directly because it's just a polynomial, which is defined for all real numbers. So the only restrictions come from the denominator. Therefore, the domain will be the set of all real numbers x where x² - 5x + 6 > 0.

Alright, so I need to solve the inequality x² - 5x + 6 > 0. Let's factor the quadratic expression. Let me try factoring x² - 5x + 6. The coefficients are 1, -5, and 6. I need two numbers that multiply to 6 and add up to -5. Hmm, factors of 6 are 1 & 6, 2 & 3. Let's see, -2 and -3 multiply to 6 and add up to -5. Yes, that works. So, x² - 5x + 6 factors into (x - 2)(x - 3). 

So, the inequality becomes (x - 2)(x - 3) > 0. Now, to solve this inequality, I can use a sign chart or test intervals. The critical points are where each factor is zero, which is x = 2 and x = 3. These points divide the real number line into three intervals: (-∞, 2), (2, 3), and (3, ∞). 

I need to test each interval to see where the product (x - 2)(x - 3) is positive. 

First interval: (-∞, 2). Let's pick a test point, say x = 1. Plugging into (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive. So this interval is part of the solution.

Second interval: (2, 3). Let's take x = 2.5. Then (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative. So this interval is not part of the solution.

Third interval: (3, ∞). Let's choose x = 4. Then (4 - 2)(4 - 3) = (2)(1) = 2, which is positive. So this interval is part of the solution.

Therefore, the solution to the inequality (x - 2)(x - 3) > 0 is x ∈ (-∞, 2) ∪ (3, ∞). 

But wait, let me double-check. When we have a quadratic inequality, the sign alternates between the intervals defined by the roots. Since the leading coefficient is positive (1), the parabola opens upwards. Therefore, the expression is positive outside the roots and negative between them. So yes, that confirms the solution is (-∞, 2) ∪ (3, ∞). 

Therefore, the domain of f(x) is all real numbers except where x is between 2 and 3, inclusive. So, the domain is (-∞, 2) ∪ (3, ∞). 

But just to make sure I didn't make a mistake in factoring. Let me check the quadratic again. x² -5x +6. Let's expand (x - 2)(x - 3) = x² -3x -2x +6 = x² -5x +6. Yes, that's correct. So the factoring is right. 

Another check: if I plug in x = 0 into the original denominator sqrt(0 -0 +6) = sqrt(6), which is fine. If x = 1, sqrt(1 -5 +6) = sqrt(2), which is fine. If x = 2, sqrt(4 -10 +6) = sqrt(0) = 0, which is not allowed. Similarly, x =3, sqrt(9 -15 +6) = sqrt(0)=0. So the endpoints 2 and 3 are excluded, which matches our solution. 

So, putting it all together, the domain is all real numbers less than 2 and greater than 3. 

Therefore, the domain of f(x) is (-∞, 2) ∪ (3, ∞). 

But wait, just to be thorough, let me check another point in each interval. For example, x = -1: denominator sqrt(1 +5 +6) = sqrt(12), which is okay. x = 2.5: denominator sqrt(6.25 -12.5 +6) = sqrt(-0.25), which is not real. x =4: sqrt(16 -20 +6) = sqrt(2), which is okay. So yes, the testing confirms the intervals where the expression inside the sqrt is positive. 

So, I think that's correct. Therefore, the domain is all real numbers except those between 2 and 3, including 2 and 3 where the denominator becomes zero. So, the answer is (-∞, 2) ∪ (3, ∞).
Therefore, the domain of the real-valued function $f(x)$ is $\boxed{(-\infty, 2) \cup (3, \infty)}$.
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