If a, B, C are real numbers, and a of B is equal to C, B of B is equal to C of a, find the value of a minus B plus C of a plus B minus C

If a, B, C are real numbers, and a of B is equal to C, B of B is equal to C of a, find the value of a minus B plus C of a plus B minus C

Let a / b = B / C = C / a = K. (1) when a + B + C ≠ 0: A / b = B / C = C / a = k, then: (a + B + C) / (B + C + a) = k, and K = 1. So a = b = C, (a + B-C) / (a-b + C) = (c + C-C) / (C-C + C) = C / C = 1; (2) when a + B + C = 0: a = - B-C; b = - a-c

1. Counter proof: non zero real numbers a, B, C constitute the arithmetic sequence with tolerance not 0, and proof: 1 / A, 1 / B, 1 / C cannot be the arithmetic sequence
2. Given ABC three real numbers, a + B + C = 0, ABC = 1
Verification: at least one of a, B, C is greater than 3 / 2

1. Suppose 1 / A, 1 / B and 1 / C are equal difference sequence, then 2 / b = 1 / A + 1 / C = (a + C) / ACB (a + C) = 2Ac, because a, B and C constitute equal difference sequence: 2B = a + C. Therefore, 2b ^ 2 = 2Ac obtains that B ^ 2 = AC is equal ratio sequence, which is inconsistent with the original equal difference sequence. 2

It is known that the positive numbers a, B and C are equal difference sequence, and the tolerance D ≠ 0. It is proved that 1a, 1b and 1C cannot be equal difference sequence

Proof (counter proof): suppose 1a, 1b and 1C are equal difference sequence, then 1b − 1A = 1c − 1b, that is, a − bab = B − CCB is multiplied by B on both sides to get a − Ba = B − CC and ∵ a, B and C are equal difference sequence, and the tolerance is not zero, ∵ A-B = B-C ≠ 0 1A = 1C.. both sides are multiplied by AC to get a = C. This is contradictory to the known sequence a, B, C whose tolerance is not zero and a ≠ C, so the sequence 1a, 1b, 1c cannot be an arithmetic sequence