FX is an odd function defined on R, XF '(x) + F (x) on (- infinity, O)

XF '(x) + F (x) = [XF (x)] & 39; & lt; 0, XF (x) on (- $, 0), XF (x) is decreasing, because the odd function is symmetric about the origin, so XF (x) also decreases at (0, + $), and f (- 2) = 0, & nbsp; so XF (- 2) = 0, we can draw a diagram of function y = XF (x) & nbsp; get the solution of XF (x) & lt; 0 at (- 2,0) U (2

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1. If f (x) is an odd function defined on R and is an increasing function in (0, + ∞), and f (2) = 0, then the solution set of the inequality XF (x) < 0 is______ ．
2. It is known that the function f (x) is a monotone increasing function defined on (0, ∞), and f (XY) = f (x) + F (y) for any X and Y in the domain of definition f（3）=1. (1) Finding the value of F (1) (2) Solving inequality f (3x) + F (2x-1) ≤ 2
3. (a+b)²－9(ac+bc)+20c² (a²+5a)²+8(a²+5a)+16
4. m^5－m^4＋m^3－2m^2＋2m－2
5. Extracting common factor and decomposing factor (x+2y)²+x+2y The polynomial A & # 178; (x-a) + a (A-X) is factorized, and the result is () A、（x-a）（a²+a） B、a（x-a）（a+1） C、a（x-a）（a-1） D、a（x+a）（a-1） 2013 times of (- 8) + 2012 times of (- 8) can be divided by () A、3 B、5 C、7 D、9 Can 23 times of 5 - 21 times of 5 be divided by 120? （a-3）²-（2a-6）
6. Factorization: - 4 (x-2y) 2 + 9 (x + y) 2
7. Do a few questions, factoring and so on X3-9x = 16x fourth power - 1 = 2x2 + 4x + 2 = 2A - (a + b) (a-b)= 3a2b(2a－3b2)= (-1＋2a)2= (a+b)2-(a-b)2= （x-2y+4）(x-2y-4)= -a2+81= ay2-2a2y+3= 9y3-2y= Calculate the value of (2-A) (a + 2) + (- b-2) (2-B), where a = √ 2, B = 2 √ 2
8. Several factorization problems 1. (1) factorization n ^ 3-N (2) if n is an integer, try to explain that n ^ 3-N is a multiple of 6 (1) (19.99 + 4.99) 2-4 * 19.99 * 4.99 Urgent need before 17:00 p.m. on December 26, thank you
9. Several factoring problems 1.（x²+2x-3)(x²+2x-24)+90 2.x^4 -4x³+4x²-9 3.(x-1)(x-2)(x-3)(x-4)-24 I hope you can give me some advice
10. 1.（3x-y)^2-3x+y 2.(2a+s)^2-6(2a+s)+9 3.4(x-y)^2+4(x-y)+1 4.(4x^2+1)^2-16x^2 5.(x-y)^3+2×(y-x)^2 6.4(3a+2x)^2-9(a-x)^2 7.y-x+3(x-y)^2
11. Function FX = 1 / 4 ^ x + m (M > 0), when X1 + x2 = 1, FX1 + FX2 = 1 / 2, 1, find the value of M 2. The known sequence an satisfies an = F0 + F (1 / N) + F (2 / N) + +F (n-1 / N), find an 3, if Sn = a1 + A2 + a3 + +An, Sn 4 ^ x + m is the denominator x1, and X2 belongs to R
12. Known function FX = x ^ 2-4x + 6, x > = 0, FX = 2x + 4, X
13. The definition of function monotonicity is as follows: if X1 is less than x2 and FX1 is less than or equal to FX2, then the function is monotone increasing function. Why is there equal to in the definition
14. For the function f (x) defined on the interval D, if ∀ x1, X2 ∈ D, and X1 & lt; X2, there is & nbsp; If f (x 1) ≥ f (x 2), then f (x) is a "non increasing function" on interval D. if f (x) is a "non increasing function" on interval [0,1] and f (0) = L, f (x) + F (l-x) = L, then f (x) ≤ - 2x + 1 holds when x ∈ [0,14]. There are the following propositions: ① ∀ x ∈ [0,1], f (x) ≥ 0; ② when x1, X2 ∈ [0,1] and x1 ≠ X2, f (x 1) ≠ f (x) ③ f (18) ）+F (511) + F (713) + F (78) = 2; 4. When x ∈ [0,14], f (f (x)) ≤ f (x). Where the serial numbers of all propositions that you think are correct are______ ．
15. If 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 points will be sent! If it's OK, points will be added! The earlier you answer, the more points will be added. Thank you
16. When x > 0, the function FX is meaningful and satisfies F 2 = 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
17. Given that f (x) is an odd function in the domain of definition on the interval [1, - 1] and is an increasing function, the value of F (0) can be obtained,
18. It is known that the function f (x) is an odd function in the domain of definition (&; - 2,2) It is known that the function f (x) is an odd function in the domain of definition (&; - 2,2) and increases monotonically in the interval [0,2]. The solution to the inequality f (x-1) + F (x ^ 2-1) < 0 The answer is (- 1,1), why
19. It is known that the function f (x) is an odd function in the domain (- 2,2). The odd function in the interval [0,2] decreases monotonically in the interval [0,2] Solve the inequality f (x-1) + F (x ^ 2-1) < 0
20. It is known that the function f (x) is an odd function whose domain of definition is R. when x > 0, f (x) = x-x + 1, then if x