Explanation:
Options B and C are wrong because 0 as a numerator is totally fine as a real solution.
In Option A,
15(x² - 1) = 10(3x - 3)
15x² - 45x + 30 = 0
x² - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 or x = 2.
This is correct as x cannot be 1 (dividing by zero).
In Option D,
(x + 2)(x - 1) = 40
x² + x - 42 = 0
(x - 6)(x + 7) = 0
x = 6 or x = -7.
In here x = 1 does not even appear in our working so it is not an extraneous solution (even though it is dividing by zero)
Hence the answer is A.