It is known that the definition field of function f (x) is all real numbers of X ≠ O. for any X1 and X2 in the definition field, f (x1 · x2) = f (x1) + F (x2), and when x > 1, f (x) > 0, f (2) = 1 (1) Verify that f (x) = f (- x) (2) F (x) is an increasing function on (0, + infinity) (3) Solving inequality f (|x| + 1) < 2

It is known that the definition field of function f (x) is all real numbers of X ≠ O. for any X1 and X2 in the definition field, f (x1 · x2) = f (x1) + F (x2), and when x > 1, f (x) > 0, f (2) = 1 (1) Verify that f (x) = f (- x) (2) F (x) is an increasing function on (0, + infinity) (3) Solving inequality f (|x| + 1) < 2

1.∵f(x ²)= f(x)+f(x)=f(-x)+f(-x)∴f(x)=f(-x)=f(x ²)／ 2. If 00, i.e. f (x) is an increasing function on (0, + infinity), 3 ∵ f (2) = 1 ∵ f [4] = f (2) × 2) = f (2) + F (2) = 2, then f (|x| + 1) < f [4] and f (x) is an increasing function on (0, + infinity)

It is known that the definition field of function f (x) is all real numbers of X ≠ O. for definition fields X1 and X2, f (x1 · x2) = f (x1) + F (x2) And when x > 1, f (x) > 0, f (2) = 1, try to compare the size of F (- 5 / 2) and f (7 / 4)

F (2 * 1) = f (2) + F (1) 1 = 1 + F (1) f (1) = 0 f [(- 1) * - 1)] = f (- 1) + F (- 1) f (1) = 2F (- 1) f (- 1) = 0 f (2 * 2) = f (2) + F (2) = 2

Let the domain of function f (x) be on the set of positive real numbers, if f (x1 + x2) = f (x1) + F (x2) and f (8) = 3 for any X1 > 0 and X2 > 0, find f (2)

F(8)=F（2+6）=F(2)+F(6)
=F(2)+F(2+4)
=F(2)+F(2)+F(4)
=F(2)+F(2)+F(2）+F(2）
=4F(2）
So f (2) = 3 / 4

It is known that for any real number x, the function f (x) satisfies f (- x) = f (x). If the equation f (x) = 0 has 2009 real solutions, the sum of these 2009 real solutions is __

Let the real solution of equation f (x) = 0 be x1, X2,..., X2009,
Let x1 ＜ x2 ＜... ＜ X2009,
And f (- x) = f (x),
If there is x0 so that f (x0) = 0, then f (- x0) = 0,
∴x1+x2009=0，x2+x2008=0，…，x1004+x1006=0，x1005=0，
∴x1+x2+…+x2009=0．
So the answer is: 0

It is known that for any real number x, the function f (x) satisfies f (- x) = f (x). If the equation f (x) = 0 has 2009 real solutions, the sum of these 2009 real solutions is __

Let the real solution of equation f (x) = 0 be x1, X2,..., X2009,
Let x1 ＜ x2 ＜... ＜ X2009,
And f (- x) = f (x),
If there is x0 so that f (x0) = 0, then f (- x0) = 0,
∴x1+x2009=0，x2+x2008=0，…，x1004+x1006=0，x1005=0，
∴x1+x2+…+x2009=0．
So the answer is: 0

The odd function f (x) defined on R satisfies: when x > 0, f (x) = 2009 ^ x + log2009 x, then the number of real roots of equation f (x) = 0 is? F (x) = 2009 ^ x + log2009 (x) is there any conversion relationship between 2009 ^ X and log2009 x

When x tends to 0 +, it means that the function tends to 0f (x) = 2010x + ㏒ 2010x2010 ^ 0 = 1 from the right. The logarithmic function is negative infinity next to 0, so f (x) 0 is only a part of it. X > 0 has a real root. When X0, f (x) = 2009x + ㏒ 2009xf (x) increases monotonically, and when x tends to 0 +, f (x) 0), so there must be one when x > 0