Given the function f (x) = log2 ((x-1) / (x + 1)), G (x) = 2aX + 1-A, and H (x) = f (x) + G (x) (1) the parity of H (x) is discussed; (2) when a = 1, it is proved that h (x) is monotonically increasing on X belonging to (1, + ∞), and that h (x) has two zeros; (3) if the equation f (x) = log2g (x) about X has two unequal real roots, find the value range of A

f(x)=log(2)[(x-1)/(x+1)],g(x)=2ax+1-a,h(x)=f(x)+g(x)
1、f(-x)=log(2)[(-x-1)/(-x+1)]=log(2)[(x+1)/(x-1)]=-log(2)[(x-1)/(x+1)]=-f(x)
G (- x) = - 2aX + 1-A, if 1-A = 0, i.e. a = 1, then G (- x) = - G (x),
If h (- x) = - f (x) - G (x) = - [f (x) + G (x)] = - H (x), then H (x) is an odd function
If a = {- log (2) [(x-1) / (x + 1)]} / (2x) = - f (x) / (2x), then G (x) = - f (x) + 1 + F (x) / (2x)
When h (x) = f (x) + G (x) = 1 + F (x) / (- 2x), H (- x) = 1 + F (- x) / (- 2x) = 1-f (x) / (- 2x) = 1 + F (x) / (- 2x) = H (x)
Then H (x) is an even function
If a takes a value other than the above two cases, then H (x) is a non odd non even function
2. When a = 1, H (x) = f (x) + G (x) = log (2) [(x-1) / (x + 1)] + 2x = log (2) (x-1) - log (2) (x + 1) + 2x
The derivation shows that H '(x) = 1 / [(x-1) LN2] - 1 / [(x + 1) LN2] + 2 = 1 / LN2 * [(x + 1-x + 1) / (x ^ 2-1)] + 2 = 2 / [LN2 * (x ^ 2-1)] + 2
X ∈ (1, + ∞), x ^ 2-1 > 0, H '(x) > 0, H (x) is a monotone increasing function on (1, + ∞)
∵ f (1) - > - ∞, ∵ H (1) - > - ∞, and H (x) is a monotone increasing function on (1, + ∞),
Therefore, H (x) must have and only have one zero point on (1, + ∞)
Moreover, when a = 1, H (x) is an odd function. From the symmetry of the odd function, we can see that h (x) must be a monotone increasing function on (- ∞, - 1)
The function H (x) has two zeros
3、f(x)=log(2)[(x-1)/(x+1)]=log(2)[g(x)] => g(x)=(x-1)/(x+1)=2ax+1-a
It is found that the equation 2aX ^ 2 + ax + 2-A = 0 has two unequal real roots
The solution of △ = a ^ 2-4 * 2A * (2-A) > 0 is a > 16 / 9 or a

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