If you plug x = 3 into choice A, then you dont have any division by zero errors or square roots of negative values.
Choice B in contrast has x-9 = 3-9 = -6 under the square root, so this leads to an imaginary/complex result. Choice D is similar.
Choice C will have zero in the denominator after plugging in x = 3 since x^2-9 = 3^2-9 = 9-9 = 0.
