It is known that f (x) is an even function, G (x) is an odd function, and f (x) + G (x) = 1 / X-1 over the common domain {x ∈ R, X ≠ ± 1}, Find the analytic expression of F (x) The correct answer is... F (x) + G (x) = 1 / (x-1)... Sorry... Ask for the correct answer again

It is known that f (x) is an even function, G (x) is an odd function, and f (x) + G (x) = 1 / X-1 over the common domain {x ∈ R, X ≠ ± 1}, Find the analytic expression of F (x) The correct answer is... F (x) + G (x) = 1 / (x-1)... Sorry... Ask for the correct answer again

When x ≠ ± 1, there is
f(x)+g(x) = 1/(x-1)
f(-x)+g(-x) = f(x)-g(x) = -1/(x+1)
The sum of the two formulas is 2F (x) = 1 / (x-1) - 1 / (x + 1) = 2 / ((x-1) (x + 1))
F (x) = 1 / ((x-1) (x + 1))

Do even functions with domain r satisfy f (0) = 0? For odd functions with domain r satisfy this condition

If the domain is an even function of R, f (0) = 0 does not necessarily hold
Because even function f (x) = f (- x), it is impossible to judge the value of F (0)
F (x) = x ^ 2 is even function, f (0) = 0
F (x) = x ^ 2-1 is also an even function, f (0) = - 1
If f (x) is an odd function, then - f (x) = f (- x)
When x = 0, - f (0) = f (0)
So f (0) = 0

It is known that the function f (x) whose domain of definition is R is even. When x ≥ 0, f (x) =, find the analytic expression of F (x), and prove that f (x) = 2 ^ (1-x) is in the interval (1,2)
F (x) = (x + 1) x has a solution on (1,2)

∵ f (x) is an even function ∵ f (x) = f (- x) ∵ x ≥ 0 ∵ - x ≤ 0 ∵ f (- x) = - X / (1-x) = x / (x-1) ∵ f (x) = x / (x + 1), (x ≥ 0) f (x) = x / (x-1), (x ≤ 0). It is proved that if∵ f (x) = 2 ^ (1-x), then 1-x ∈ R ∵ x ∈ R ∵ f (x) = 2 ^ (1-x) has a solution in the interval (1,2)