Let f (x) be a function defined on R. for any x, y ∈ R, there is always f (x + y) = f (x) · f (y). When x > 0, there is 0

Let f (x) be a function defined on R. for any x, y ∈ R, there is always f (x + y) = f (x) · f (y). When x > 0, there is 0

(1) Let x = 0, y = 0, so f (0) = f ^ 2 (0), f (0) [f (0) - 1] = 0, so f (0) = 0
F (0) = 0 or F (0) = 1. When f (0) = 0, for x > 0, f (x) = f (x + 0) = f (x) f (0) = 0, and when x > 0, there is 0

Let f (x) be a function defined on a positive integer set and satisfy f (x + 2) = f (x + 1) - f (x). If f (1) = Lg3 / 2, f (2) = LG15, find the value of F (2007)

Because f (x + 2) = f (x + 1) - f (x) f (x + 3) = f (x + 2) - f (x + 1) = f (x + 1) - f (x) - f (x + 1) = - f (x) f (x + 6) = - f (x + 3) = f (x), f (x) is a function with a period of 6, because f (2007) = f (6 * 334 + 3) = f (3) because f (3) = f (2) - F (1) = lg15-lg3 / 2 = LG10 = 1

Let f (x) be a function defined on R and satisfy f (x + 2) = f (x + 1) - f (x). If f (1) = LG32, f (2) = LG15, then f = 1___ ．

（1）f（1）=lg32，f（2）=lg15，∴f（3）=f（2）-f（1）=lg15-（lg3-lg2）=lg5+lg2=1，f（4）=f（3）-f（2）=1-lg15，f（5）=f（4）-f（3）=1-lg15-1=-lg15，f（6）=f（5）-f（4）=-lg15-（1-lg15）=-1，f（7）=f（...

If a is a set of some real numbers, and 0 does not belong to a, 1 does not belong to a, if a ∈ a, then 1 / 1-A ∈ a,
(1) How many elements are there in any given real number a, a ≠ 0, a ≠ 1?
∵ when a ≠ 0, a ≠ 1, 1 / [1 - (1 / 1-A)] = A-1 / A, 1 / [1 - (A-1 / a)] = a, and a ≠ (1 / 1-A) ≠ (A-1 / a), ∵,
There are three elements in a: A, 1 / 1-A, A-1 / A
I don't know 1 / [1 - (1 / 1-A)] = A-1 / A, 1 / [1 - (A-1 / a)] = a, and a ≠ (1 / 1-A) ≠ (A-1 / a),

A ∈ a, then 1 / 1-A ∈ a, which means that any element X in a, take it into 1 / 1-x, as long as it is not equal, the resulting value also belongs to A1 / [1 - (1 / 1-A)] = A-1 / A. This formula is to take x = 1 / 1-A into 1 / 1-x, 1 / [1 - (A-1 / a)] = a, this formula is to take x = A-1 / a into 1 / 1-x, and then go down to cycle, so there are three

If a belongs to a, then 1 / 1-A belongs to a
(1) If 2 belongs to a, find a
(2) Can set a be a single element set? If so, find out a. no, explain the reason
(3) Verify that 1-1 / a belongs to a
If you pull Baidu, I need some detailed answers.

If a ∈ a, then 1 / 1-A ∈ a (1) ∵ 2 ∈ A. according to the condition ∵ 1 / 1-2 ∈ a, that is - 1 ∈ A. and ∵ - 1 ∈ a, that is 1 / 1 - (- 1) ∈ a, that is 1 / 2 ∈ a

If a denotes a real number, is 1 / a a real number?

Not necessarily, a cannot be 0

a. B, C are real numbers, and a * B + b * C + C * a = 1, then the following holds true:
A a^2+b^2+c^2 >=2
B (a+b+c)^2 >=3
C 1 / A + 1 / B + 1 / C > = Double radical three
D a+b+c

What about C and D?

Let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ A. There are at least three elements in the set a

For reference only: as shown in the figure
 
 

Let the elements in the set a be real numbers and satisfy the following conditions: there is no 1 in A. if a ∈ a, then there must be 1 / (1-A) ∈ a
It is proved that if 2 ∈ a, there must be two other elements in a, and these two elements can be obtained

From "a ∈ a, there must be 1 / (1-A) ∈ a" and "2 ∈ a", we deduce 1 / (1-2) = - 1 ∈ A. similarly, 1 / [1 - (- 1)] = 1 / 2 ∈ A. here - 1 and 1 / 2 are the two elements

Let the elements of s in the set be real numbers and satisfy the following conditions: (1) there is no number 1 in S. (2) if a belongs to s, then 1 / 1-A must belong to s
1. Prove that if 2 belongs to s, then there must be two elements in s, and find out these two elements,
Above are the conditions, below are the problems,
-- yes, I'll give it to you

If a = 2,1 / 1-A = - 1
If a = - 1,1 / 1-A = 1 / 2
If a = 1 / 2, 1 / 1-A = 2
That is, if 2 belongs to s, there must be two other elements in the set
-1,1/2