If the domain of F (x), G (x) is r, f (x) is odd function, G (x) is even function, and f (x) + G (x) = 1 / (x ^ 2 + X + 1), then f (x)=

f(x)+g(x)=1/（x^2+x+1)
f(-x)+g(-x)=1/(x^2-x+1)
That is (with parity)
-f(x)+g(x)=-1/(x^2-x+1)
So (subtracting two equations)
f(x)=1/2[1/(x^2+x+1)-1/(x^2-x+1)]

Let f (x) and G (x) be defined in such a way that {x L x belongs to R and X is not equal to plus or minus one} f (x) is even function and G (x) is odd function And f (x) + G (x) = 1 / (x-1), find the analytic formula of F (x) and G (x)

F （x）+g（x）=1/（x-1） 1
F （-x）+g（-x）=1/（-x-1） 2
Because f (x) is even function. G (x) odd function
So f (- x) = f (x), G (- x) = - G (x)
So formula 2 becomes f (x) - G (x) = 1 / (- x-1) 3
1 + 3, with 2F (x) = 1 / (- x-1) + 1 / (x-1) = [(x-1) + (- x-1)] / (1-x?) = 2 / (x-1)
So f (x) = 1 / (x? - 1)
Substituting into 1, G (x) = x / (x? - 1)

It is known that f (x) is an even function on R, G (x) is an odd function on R, and G (x) = f (x-1). If G (- 1) = 2, then the value of F (2008) is () A. 2 B. 0 C. -2 D. ±2

∵ f (x) is an even function on R, G (x) is an odd function on R ? f (- x) = f (x), G (- x) = - G (x) ? g (x) = f (x-1) ? g (- x-1) = f (- x-1) = - G (x) ? g (x) = - f (- x-1) = f (x-1) so that X-1 = t, then x = 1 + T  - f (T) = f (- T-2), that is, f (T + 2) = - f (t) ? f (x) = f (x-1)

Given that f (x-1) is an odd function, f (x) is an even function, f (2008) = 1, then f (4) =?

f(x+1)=f(-x+1)=f(-(x-2)-1)=f(x-2-1)=f(x-3)
f(x)=f(x-4)
The function has a period of 4
f(2008)=1

If the odd function f (x) defined on R satisfies f (x + 2) = - f (x), then the value of F (2008) is () A. -1 B. 1 C. 2 D. 0

From F (x + 2) = - f (x), f (x + 4) = - f (x + 2) = - [- f (x)] = f (x),
That is, the period of the function f (x) is 4
So f (2008) = f (502 × 4) = f (0)
Because f (x) is an odd function defined on R,
Therefore, according to the properties of odd functions, we know that f (0) = 0,
So f (2008) = 0
Therefore, D

Let f (x) be an odd function defined in R, f (x + 1) = - f (x). When x belongs to [0,1], f (x) = x, then f (2008) =? Why is 0 not - 1

Because the function f (x) is an odd function defined in R
So there is f (x) = - f (x)
Because the function satisfies f (x + 1) = - f (x)
So f (x) = f (x + 1)
And f (x) = x (holds when x belongs to [0,1])
So f (0) = 0
Then there is f (1) = f (0) = 0; f (2) = f (1) = 0; f (3) = f (2) = 0
And so on: F (2008) = 0

Let f (x) be an odd function defined on R. if f (x + 3) × f (x) = - 1, f (- 1) = 2, then f (2008) = -___ ?

∵f(x+3)*f(x)=-1 .(1)
∴f(-x+3)*f(-x)=-1
f(3-x)f(x)=1 .(2)
The formula (1) + (2) is obtained
[f(x+3)+f(3-x)]f(x)=0
∵ f (x) is not always 0
∴f(x+3)+f(3-x)=0
That is, f (x + 3) = - f (3-x) = f (x-3)
f(x)=f(x+6)
That is, f (x) is a periodic function with a period of 6
∴f(2008)=f(6*334+4)=f(4)=

Let f (x) be an odd function defined on R, and f (x + 3) × f (x) = - 1, f (- 1) = 2, then f (2008) = () A0.5 B0 C2 D-1

Given that the function f (x) defined on R is odd, and the period of function f (3x + 1) is 3 and f (1) = 5, then the value of F (2007) + F (2008) is () A. 0 B. 5 C. 2 D. -5

∵ the period of the function f (3x + 1) is 3,  f [3 (x + 3) + 1] = f (3x + 1), that is, f [(3x + 1) + 9] = f (3x + 1), f (x) is a function with period 9; and f (x) is an odd function on R,

Given that the function f (x) defined on R is odd, and the period of function f (3x + 1) is 3 and f (1) = 5, then the value of F (2007) + F (2008) is () A. 0 B. 5 C. 2 D. -5

∵ the period of the function f (3x + 1) is 3,  f [3 (x + 3) + 1] = f (3x + 1), that is, f [(3x + 1) + 9] = f (3x + 1), f (x) is a function with period 9; and f (x) is an odd function on R,

If the domain of F (x), G (x) is r, f (x) is odd function, G (x) is even function, and f (x) + G (x) = 1 / (x ^ 2 + X + 1), then f (x)=