Simplify. \[ \frac{\frac{12}{x-7}+x}{\frac{3}{x-7}+1} \]

Let's simplify the expression \[ \frac{\frac{12}{x-7}+x}{\frac{3}{x-7}+1}. \] Start by simplifying the numerator \(\frac{12}{x-7}+x\). To do this, we can express \(x\) with a common denominator of \(x-7\): \[ x = \frac{x(x-7)}{x-7} = \frac{x^2 - 7x}{x-7}. \] Now the numerator becomes \[ \frac{12}{x-7}+x = \frac{12}{x-7} + \frac{x^2 - 7x}{x-7} = \frac{12 + x^2 - 7x}{x-7}. \] Thus, the numerator simplifies to: \[ \frac{x^2 - 7x + 12}{x-7}. \] Next, consider the denominator \(\frac{3}{x-7}+1\). We express \(1\) with the common denominator \(x-7\): \[ 1 = \frac{x-7}{x-7}. \] So the denominator becomes \[ \frac{3}{x-7} + 1 = \frac{3}{x-7} + \frac{x-7}{x-7} = \frac{3 + (x-7)}{x-7} = \frac{x - 4}{x-7}. \] Now substituting both simplified parts back into the original expression gives: \[ \frac{\frac{x^2 - 7x + 12}{x-7}}{\frac{x - 4}{x-7}}. \] We can simplify this fraction by multiplying by the reciprocal of the denominator: \[ \frac{x^2 - 7x + 12}{x-7} \cdot \frac{x-7}{x-4} = \frac{x^2 - 7x + 12}{x - 4}. \] Notice that the \(x-7\) terms cancels out as long as \(x \neq 7\). The remaining task is to factor the numerator \(x^2 - 7x + 12\). We can factor it as follows: \[ x^2 - 7x + 12 = (x - 3)(x - 4). \] Thus, we can rewrite our expression as: \[ \frac{(x-3)(x-4)}{x-4}. \] Now we can cancel the \(x-4\) term from the numerator and the denominator, provided \(x \neq 4\): \[ x - 3. \] Thus, the simplified expression is \[ \boxed{x - 3} \] for \(x \neq 4\) and \(x \neq 7\).