Given that f (x) is an increasing function defined on (0, + ∞), and satisfies f (XY) = f (x) + F (y), f (2) = 1, find the solution set of the inequality f (x) - f (X-2) > 3

Given that f (x) is an increasing function defined on (0, + ∞), and satisfies f (XY) = f (x) + F (y), f (2) = 1, find the solution set of the inequality f (x) - f (X-2) > 3

So f (2) = f (2) + F (1) = 4
F (x) - f (X-2) > 3, that is, f (x) > 3 + F (X-2) = f (8) + F (X-2) = f (8x-16), because f (x) is an increasing function defined on (0, + ∞)
So x > 8x-16, so 0
Set u = {x | x ≤ 10, and X ∈ n *}, a is really contained in U, B is really contained in U, and a ∩ B = {3,5}, (cub) ∩ a = {1,2,4}, (CUA)
Set u = {x | x ≤ 10, and X ∈ n *}, a is really contained in U, B is really contained in U, and a ∩ B = {3,5}, (cub) ∩ a = {1,2,4}, (CUA) ∩ (cub) = {6,7} find sets a and B
Let's use a simple graphic method
The {} in the set a represents the {} in the set B represents the U
Then ([1,2,4 {3,5] 8,9} 6,7)
So a = {1,2,3,4,5}, B = {3,5,8,9}
A(1,2,3,4,5)B(3,5,8,9)
What does cub mean? Please explain
Why do we need to verify Δ with Veda theorem
Given that f (x) is an increasing function defined on (0, + ∞), the solution set of the inequality f (x) > F (8x-16) is ()
A. （0，+∞）B. （0，2）C. （0，167）D. （2，167）
From the meaning of the title, we can get x ＞ 8x-16 ＞ 0 and 2 ＜ x ＜ 167
Let u = {x belongs to positive integer / X ≤ 8}, if a ∩ (cub) = {2.8}, (CUA) U (cub) = {1.2.3.4.5.6.7.8}, find the set a
Why is the answer to this question a = {2.8}. Why is there no other element? Is it because I don't understand the concept or what? It's better to have steps of disintegration
Because u = {x ∈ n + | x ≤ 8}, so
U={1,2,3,4,5,6,7,8},
Because a ∩ (cub) = {2,8},
So there are 2,8 in a and no 2,8 in B
(CUA) ∪ (cub) = Cu (a ∩ b)
Cu (a ∩ b) = {1,2,3,4,5,6,7,8}, which means that a ∩ B is an empty set. It means that there is no repeating element in ab
Suppose there is 1 in a, then there is no 1 in B, then there is 1 in cub
It is in contradiction with a ∩ (cub) = {2,8}, so there is no 1 in a, which can be deduced in the same way
34567 does not belong to a,
So the set a is {2,8}
How to prove the generalization of Weida's theorem
Be more detailed
Let x1, X2 , xn are n solutions of the equation ∑ AIX ^ i = 0
There are: an (x-x1) (x-x2) (x-xn)=0
So: an (x-x1) (x-x2) (x-xn) = ∑ AIX ^ I (on (x-x1) (x-x2) (x-xn) is better to use the principle of multiplication.)
Through coefficient comparison, it can be concluded that:
A（n-1）=-An(∑xi)
A（n-2）=An(∑xixj)
...
A0==(-1)^n*An*∏Xi
So: Σ xi = (- 1) ^ 1 * a (n-1) / a (n)
∑XiXj=(-1)^2*A(n-2)/A(n)
...
∏Xi=(-1)^n*A(0)/A(n)
Where ∑ is the sum and Π is the product
According to the spherical root formula x = (- B ± △) (△ = the square of B - 4 * a * c), the answer can be obtained by multiplying the positive and negative cases.
F (x) is the inequality f (2x-1) of decreasing function solution defined on [- 1,2]
The problem is equivalent to a system of inequalities
2 x - 1 >= -1
2 x - 1 = -1
1 - 2 x 1 - 2x
The solution is 1 / 2 < X
-1≤2x-1≤2
-1≤1 - 2x ≤2
2x-1＞1-2x
If u = R, a = (x | X. > equals 0), B = (Y | Y > 1), then what is the relationship between CUA and cub?
CuA={X|X＜0} ,CuB={Y|Y≤1}
10. Y are all real numbers that conform to the set, but the notation is different. In essence, they are all real numbers. Therefore, CUA is included in cub
Because u = R, a = (x | X. > equals 0), B = (Y | Y > 1),
So CuX is a
The proof method of Vader's theorem!
In this paper, we discuss the Veda theorem of degree 2
I have learned two kinds. One is to prove it by the root formula. The other is to divide both sides of ax ^ 2 + BX + C = 0 by a, then expand it by (x-x1) (x-x2) = 0, and then compare the two formulas
I believe there must be many ways ~ I hope you can provide more proof methods~
If the function FX satisfies: for all real numbers x and y, FX + FY = x (2y-1) holds. (1) find F0
Let x = 0, y = 0, then F0 + F0 = 0, 2f0 = 0, F0 = 0. Let x = 1, y = 0, then F1 + F0 = - 1, F1 = - 1