The function y = f (x) defined on positive integer set has f (a + b) = f (a) * f (b) constant for any a, B ∈ n Let f (1) = a ≠ 0, if an = f (n) (n ∈ n +) (1) Prove: the sequence {an} is an equal ratio sequence, and find out the general term formula of the sequence {an} (2) If Sn = a1 + A2 + +An, the sequence {sn-2an} is an equal ratio sequence, find the value of real number a

The function y = f (x) defined on positive integer set has f (a + b) = f (a) * f (b) constant for any a, B ∈ n Let f (1) = a ≠ 0, if an = f (n) (n ∈ n +) (1) Prove: the sequence {an} is an equal ratio sequence, and find out the general term formula of the sequence {an} (2) If Sn = a1 + A2 + +An, the sequence {sn-2an} is an equal ratio sequence, find the value of real number a

(1) Let's know that, f (2) = f (2) (f (2) = f (1) = f (1) (f (2) = f (1), and when n ≥ 2, an = f (n) = f (n (n-1) + 1] = f (n-1 (n-1) + 1 (n-1) + 1] = f (n-1) (f (n-1-1) + (n-1,2,...) (2) when n ≥ 2, an = f (n = 2) when n ≥ 2, an = f (n = 2) when n ≥ 2, an = n (n = 2) is n = 2, and it is obvious that when a = 1, an = 1, an n = 1, Sn = 1, sn-2n-2a, sn-2a, sn-2a, sn-2a, and so let a ≠1. Let BN = BN = sn-2an, B1 = sn-2an, B1 = -2an, B1 = -2a, B1 = -a, B2 = -a, B2, B2 = -a according to the problem set, [a (1-A)] & sup2; = A & sup2; (A & sup2; - A-1) = = = = > A = 2

It is known that the function f (x) is defined on the set of natural numbers, and for any X that belongs to a positive integer, the description is: there is f (x)

F (x) = f (x-1) + F (x + 1), that is: F (x-1) = f (X-2) + F (x) two formula sum, get: F (x + 1) + F (X-2) = 0, that is, f (X-2) = - F (x + 1), that is, f (x) = - f (x + 3) f (x + 3) = - f (x + 6), so f (x) is a periodic function, the minimum positive period is 62008 / 6 = 334... 4f (2008) = f (4) = f (1

The function f is defined on the set of positive integers, f (1) = 1, f (3) = 3, and for every positive integer n, f (2n) = f (n), f (4N + 1) = f (2n + 1) - f (n), try to determine the value of 2000
rt
Try to determine the number of all positive integers n with F (n) = n within 2000
F (1) = 1, f (3) = 3 I am not very clear, which session of the IMO seems to be the question
It can be solved in binary system

Well I feel that there are few conditions. We can't find the value of F (4N + 3), such as f (7)
Is this the condition given by the landlord? There is no rule in it Or did the landlord ask for it? But I couldn't ask for it
Ah, I can't find out. I suggest the landlord raise the reward and re issue the question

X & # 178; + 2012x-2012 = 0 use the cross decomposition method of factorization to solve the quadratic equation of one variable
x²-2012x-2012=0

x^2-2011x-2012=0
(x+1)(x-2012)=0
So x = - 1 or x = 2012

1： (x + 2) (x + 3) + the square of x minus 4
2： The square of (n + 5) minus the square of (n-1)
3： The square of (AX + by) minus the square of (BX + ay)

1、 The original formula = (x + 2) (x + 3) + (X & # 178; - 4) = (x + 2) (x + 3) + (x + 2) (X-2) = (x + 2) (x + 3 + X-2) = (x + 2) (2X + 1) 2. The original formula = (n + 5) & # 178; - (n-1) & # 178; = (n + 5 + n-1) (n + 5-N + 1) = 12 (n + 2) 3. The original formula = (AX + by) & # 178; - (BX + ay) & # 17

(x+y)²+4(x+y)-5
（2a+b)²-4(2a+b)+3
6y²-11xy-10x²
ax²-a³-2ax——2a²

(x+y)²+4(x+y)-5
=（x+y+5)(x+y-1)
(2a+b)²-4(2a+b)+3
=(2a+b-3)(2a+b-1)
6y²-11xy-10x²
=(2y-5x)(3y+2x)
ax²-a³-2ax——2a²
=a(x²-a²)-2a(x+a)
=a(x+a)(x-a)-2a(x+a)
=a(x+a)(x-a-2)

(1) Factorization of 6A ^ 2 + ab-2b ^ 2-A + 11b-15
(2) Factorization of 1 + A + a (1 + a) + a (1 + a) ^ 2 + a (1 + a) ^ 3

1. If x ^ 2 (x + 1) + y (XY + y) = (x + 1) · a, then a=____ ?
2. Given X-Y = 1, xy = 2, then the value of x ^ 3-2x ^ 2Y ^ 2 + XY ^ 3 is_____ ?
3.169(a+b)^2-121(a-b)^2
4.(x^2-2x)^2-14(x^2-2x)-15
5.(x+2)(x+4)+1
That's all. We have to analyze the steps

1.A×(x+1)=(x^2+y^2)(x+1)
A=x^2+y^2
2. The original formula = XY (x ^ 2-2xy + y ^ 2)
=xy(x-y)^2
=2×1
=2
3. The original formula = (13a + 13b) ^ 2 - (11a-11b) ^ 2
=(13a+13b-11a+11b)(13a+13b+11a-11b)
=(2a+24b)(24a+2b)
=4(a+12b)(12a+b)
4. Let x ^ 2-2x = a
Original formula = a ^ 2-14a-15
=(a-15)(a+1)
=(x^2-2x-15)(x^2-2x+1)
=(x-5)(x+3)(x-1)^2
5. The original formula = x ^ 2 + 6x + 8 + 1
=x^2+6x+9
=(x+3)^2
Add points! I'm so tired!

On the practice of "integral multiplication and division and factorization" in grade two of junior high school
Proof: for any number x, y, the integral X & sup2; + 2x + Y & sup2; - 6y + 12

（x+1）²+（y-3）² +2

Given 1 + X + x2 + X3 + X4 = 0, find 1 + X + x2 + X3 + +The value of X2009______ ．

1+x+x2+x3+… +x2009=（1+x+x2+x3+x4）+（x5+x6+… +x9）+… +（x2005+x2006+… +x2009）=（1+x+x2+x3+x4）+x5（1+x+x2+x3+x4）+… +x2005（1+x+x2+x3+x4）=（1+x+x2+x3+x4）（1+x5+x10+… +Since 1 + X + x2 + X3 + X4 = 0, the original formula is 0