Let the definition field of function FX be r and satisfy the following conditions: there exists x1 ≠ x2 such that FX1 ≠ FX2 holds for any x, y, FX + y = fxfy. Find F0

Let the definition field of function FX be r and satisfy the following conditions: there exists x1 ≠ x2 such that FX1 ≠ FX2 holds for any x, y, FX + y = fxfy. Find F0

 
Given the set a = {P | x2 + 2 (p-1) x + 1 = 0, X ∈ r}, find the set B = {y | y = 2x-1, X ∈ a}
If x2 + 2 (p-1) x + 1 = 0 has roots, then △ = 4 (p-1) 2-4 ≥ 0, the solution is p ∈ (- ∞, 0] ∪ [2, + ∞), that is, set a = (- ∞, 0] ∪ [2, + ∞), so set B = {y | y = 2x-1, X ∈ a} = (- ∞, - 1] ∪ [3, + ∞)
Factorization of quadratic equation of one variable in grade nine
4X＾-45＝31X
（X＋8）（X＋1）＝-12
（3X+2）（X＋3）＝X+14
＾ denotes the quadratic power,
-3X^+22X-24=0
2(X+3)^=X(X+3)
4x ＾ - 45 = 31x 4x ＾ - 31x - 45 = 0 (4x + 5) (X-9) = 0, so x = - 5 / 4 or x = 9 (x + 8) (x + 1) = - 12, the square of X + X + 8x + 8 + 12 = 0, the square of X + 9x + 20 = 0 (x + 5) (x + 4) = 0, so x = - 5 or x = - 4 (3x + 2) (x + 3) = x + 14, the square of 3x + 10x-8 = 0 (3x-2) (x + 4
If the function f (x) defined on R satisfies: F (x1 + x2) = f (x1) + F (x2) + 1 for any x1, X2 ∈ R, then the following statement must be correct ()
A. F (x) - 1 is odd function B. f (x) - 1 is even function C. f (x) + 1 is odd function D. f (x) + 1 is even function
∵ for any x1, X2 ∈ R, f (x1 + x2) = f (x1) + F (x2) + 1, let X1 = x2 = 0, f (0) = - 1, let X1 = x, X2 = - x, f (0) = f (x) + F (- x) + 1, Let f (x) + 1 = - f (- x) - 1 = - [f (- x) + 1], and let f (x) + 1 be an odd function
Given the set a = {P | x2 + 2 (p-1) x + 1 = 0, X ∈ r}, find the set B = {y | y = 2x-1, X ∈ a}
If x2 + 2 (p-1) x + 1 = 0 has roots, then △ = 4 (p-1) 2-4 ≥ 0, the solution is p ∈ (- ∞, 0] ∪ [2, + ∞), that is, set a = (- ∞, 0] ∪ [2, + ∞), so set B = {y | y = 2x-1, X ∈ a} = (- ∞, - 1] ∪ [3, + ∞)
Mathematical problems of quadratic equation of two variables
Suppose there are x people in total, then there are (x / 3 + 2) books in total, and (X-9) / 2 books in total
X / 3 + 2 = (X-9) / 2, multiply both sides by 6 at the same time
2X+12=3(X-9)
2X+12=3X-27
3X-2X=12+27
X=39
A: there are 39 people and 15 books in total. Can you transform the equation of first degree into the equation of second degree?
Yes, if there are y books in total, there will be a system of equations
Y=X/3+2
Y=(X-9)/2
X/3+2=(X-9)/2
The same as above, the solution is x = 39, y = 15
There are x people and Y books
If the function FX defined on R belongs to R for any x1.x2, f (x1 + x2) = FX1 + fx2-1 holds, and if x > 0, FX > 1
1. Prove that GX = FX-1 is an odd function
2. Prove that FX is an increasing function on R
3. If f (4) = 5, solve the inequality f (3m ^ 2-m-2)
1. Prove: let x2 = 0, then f (x) = f (x) + F (0) - 1, then f (0) = 1
Then, f (0) = f (x + (- x)) = f (x) + F (- x) - 1, where [f (x) - 1] + [f (- x) - 1] = 0, that is, G (x) + G (- x) = 0 holds for any x ∈ R
Therefore, G (x) = f (x) - 1 is an odd function
2. It is proved that if X1 > x2 and x1-x2 = Δ x > 0, then f (x1) - f (x2) = f (x2 + Δ x) - f (x2) = f (x2) + F (Δ x) - 1-f (x2) = f (Δ x) - 1
Because when x > 0, f (x) > 1, so f (Δ x) - 1 > 0 is constant, so f (x1) > F (x2)
So, f (x) is an increasing function of R
F (4) = 2F (2) - 1 = 5, f (2) = 3, so f (3m & # 178; - m-2)
Did you type the wrong question? Isn't the minus sign in the middle??? Question: is it a minus sign
Set M = {y | y = x2-2x-1, X ∈ r} = {y | y = (x-1) 2-2, X ∈ r} = {y | y ≥ - 2} = {x | x ≥ - 2}
Which master will explain why m is equal to {x | x ≥ - 2} in the end
I would appreciate it if I could explain it clearly
Also, I'm a rookie in mathematics, and the answer should be easy to understand. Thank you (*^__ ^*) .
Set M = {y | y = x & # 178; - 2x-1, X ∈ r}
Is the range of quadratic function y = x & # 178; - 2x-1, X ∈ R
The formula is y = x & # 178; - 2x-1 = (x-1) &# 178; - 2
When x = 1, the minimum value of Y is - 2
∴M={y|y≥-2}
In set description, X and y are only real numbers
It's just a symbol,
∴M={y|y≥-2}=｛x|x≥-2}
This is about to talk about the understanding of set. Set is made up of elements. You should know that your topic is to use the description method of set to represent set. In this representation method, how to judge what elements are? It is y in {y | y = x2-2x-1, X ∈ r} or X in {x | x ≥ - 2}. For the set {y | y ≥ - 2}, which means that the set element is y, what does Y mean? Is a real number greater than or equal to - 2, and in the set {x | x ≥ - 2}, the element is x, and what does x mean? It's also a real number greater than or equal to - 2. That is to say, these two sets represent the expansion of
This is about to talk about the understanding of set. Set is made up of elements. You should know that your topic is to use the description method of set to represent set. In this representation method, how to judge what elements are? It is y in {y | y = x2-2x-1, X ∈ r} or X in {x | x ≥ - 2}. For the set {y | y ≥ - 2}, which means that the set element is y, what does Y mean? Is a real number greater than or equal to - 2, and in the set {x | x ≥ - 2}, the element is x, and what does x mean? It's also a real number greater than or equal to - 2. That is to say, the elements represented by these two sets are the same, they are all real numbers greater than or equal to - 2, but the two sets are represented by different letters! ha-ha. So don't worry about the letters. Using X is just one of our habits. Put it away
remember
{y|y=x2-2x-1,x∈R} ①
{y|y=(x-1)2-2,x∈R} ②
{y|y≥-2} ③
{x|x≥-2} ④
It can be seen from ③ y ≥ - 2 and ② y = (x-1) 2-2 that: (x-1) 2-2 ≥ - 2
But the solution is x ∈ R
To do a set problem, you first need to see what the element is. The description method is that the element in front of the vertical line represents the element.
The {y | y = x2-2x-1, X ∈ r} formula {= {y | y = (x-1) ^ 2-2, X ∈ r} y = (x-1) ^ 2-2, X ∈ R represents the number set of Y > = - 2.
This set {x | x ≥ - 2} also represents the set of numbers greater than or equal to - 2.
They represent the same set, so m is finally equal to {x | x ≥ - 2}... Expansion
To do a set problem, you first need to see what the element is. The description method is that the element in front of the vertical line represents the element.
The {y | y = x2-2x-1, X ∈ r} formula {= {y | y = (x-1) ^ 2-2, X ∈ r} y = (x-1) ^ 2-2, X ∈ R represents the number set of Y > = - 2.
This set {x | x ≥ - 2} also represents the set of numbers greater than or equal to - 2.
They represent the same set, so m is finally equal to {x | x ≥ - 2}
The first one is square x, please
Let x 2 - 2x - 1 ≥ - 2, then x belongs to all real numbers
Let (x-1) 2-2 ≥ - 2 get x ≥ 1
That's the first three solutions
{y|y=x2-2x-1,x∈R}={y|y=(x-1)2-2,x∈R}
Explanation: the formula is y = x ^ 2-2x-1 = (x-1) ^ 2-2.
{y|y=(x-1)2-2,x∈R}={y|y≥-2}
Explanation: since the image of quadratic function y = x ^ 2-2x-1 is a parabola with vertex (1, - 2), y ≥ - 2 is obtained.
{y|y≥-2}={x|x≥-2}.
Explanation: the number set {y | y... Is expanded
{y|y=x2-2x-1,x∈R}={y|y=(x-1)2-2,x∈R}
Explanation: the formula is y = x ^ 2-2x-1 = (x-1) ^ 2-2.
{y|y=(x-1)2-2,x∈R}={y|y≥-2}
Explanation: since the image of quadratic function y = x ^ 2-2x-1 is a parabola with vertex (1, - 2), y ≥ - 2 is obtained.
{y|y≥-2}={x|x≥-2}.
Explanation: both {y | y ≥ - 2} and {x | x ≥ - 2} indicate that the numbers in the set are not less than - 2.
Explanation: put it away
Known about X, y equations {x + y = 2010 solution can make the equation X-Y = 2009 hold, find M
{4x+2y=3m+2007
The value of
{4x+2y=3m+2007
2(2X+y)=3m+2007
x-y=2009
X=2009+Y
2009+Y+Y=2010
2Y=1
Y=1/2
X=4019/2
8038+1=3m+2007
6032=3m
m=6032/3
X + y = 2010 (1), X-Y = 2009 (2). (1) formula + (2) formula, 2x = 4019. x=2009.5，y=0.5
4x+2y=8038+1=8039,3m=8039-2007=6032。 m=6032/3=2010.666666。。。。。
If FX / X is a decreasing function, f (x1 + x2) is proved
Prove: because x1, X2 position is equal, might as well arrange an order for them, suppose x1