Fxy = FX + FY, F1 / 3 = 1, if FX + f2-x < 2, find the value range of X Let y = FX be the subtractive function over the definition field R

Fxy = FX + FY, F1 / 3 = 1, if FX + f2-x < 2, find the value range of X Let y = FX be the subtractive function over the definition field R

Because f (XY) = f (x) + F (y),
So f (1 / 9) = f (1 / 3) + F (1 / 3) = 2
Thus, the inequality f (x) + F (2-x) < 2
It can be reduced to f [x (2-x)] and f (x) is a subtractive function, so
x(2-x)>1/9
9x ²- 18x+1<0
The solution is (3-2 √ 2) / 3

XY belongs to RFX + y = FX + FY. Verify that FX is an odd function

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When x > 0, the function FX is meaningful and satisfies F2 = 1, f (XY) = FX + FY, FX is an increasing function (1) Verification: F (1) = 0 (2) If f (3) + F (4-8x) > 2, find the value range of X

（1 ）
f(xy)=fx+fy
Let x = 2, y = 1. Get f (1 * 2) = f (2) + F (1), so f (1) = 0
(2) Use that equation
f(3)+f(4-8x)=f(3+4-8x)>f(2*2)=f(2)+f(2)=2
Because FX is an increasing function, 3 + 4-8x > 2 * 2 is X

The function f (x) defined in (- 1,1) satisfies 1. For any x, y belongs to (- 1,1), there is f (x) + F (y) = f ((x + y) / (1 + XY)) 2. When x belongs to（ 2. When x belongs to (- 1,0), there is f (x) > 0. The first question judges the parity of FX in (- 1,1), (2) judges the monotonicity of FX in (- 1,1), (3) verifies f (1 / N) ²+ 3N + 1) = f (1 / N + 1) - f (1 / N + 2) (n belongs to n positive]

1) Let x = y = 0, from F (x) + F (y) = f ((x + y) / (1 + XY)), f (0) = 0 is obtained
Let y = - x, then there is f (x) + F (- x) = f (0) = 0, that is, f (- x) = - f (x)
So the function f (x) is an odd function
2) Set-1

The function FX defined on R satisfies f (x + y) = f (x) + F (y) + 2XY [XY belongs to R] f (1) = 2, then f (- 2) =? Please be more detailed! 20 minutes! OK, plus points! The sooner you answer, the more. Thank you

Let x = - 2 y = 1 f (- 2 + 1) = f (- 2) + F (1) - 4 so f (- 1) = f (- 2) - 2 let x = - 1 y = 1 f (- 1 + 1) = f (- 1) + f (1) - 2 so f (0) = f (- 1) let x = 0 y = 1 f (0 + 1) = f (0) + F (1) so f (0) = 0 to sum up, f (- 2) = f (- 1) + 2 = f (0) + 2 = 2

1. The function y = FX is always (fx1-fx2) / x1-x2 > 0 for any real number X1 and X2 (x1 is not equal to x2) in the definition field, then the function y = FX is In the definition domain is a monotone increasing function B monotone decreasing function C Changshu function D is not a monotone function

That is, when x1-x2 > 0, f (x1) - f (x) > 0
When x1-x2 < 0, f (x1) - f (x) < 0
So increasing
Choose a