Let f (x) be defined as R, if f (π / 2) = 0, f (π) = - 1 Let f (x) be defined as R, if f (π / 2) = 0, f (π) = - 1, and f (x1) + F (x2) = 2F (x1 + x2 / 2) f (x1-x2 / 2) (1) find f (0) (2) prove that f (x) is even function, and f (π - x) = - f (x) (3) if - π / 2 < x < π / 2, f (x) > 0, prove that f (x) monotonically decreases on [0, π]_ ∩)O~

Let f (x) be defined as R, if f (π / 2) = 0, f (π) = - 1 Let f (x) be defined as R, if f (π / 2) = 0, f (π) = - 1, and f (x1) + F (x2) = 2F (x1 + x2 / 2) f (x1-x2 / 2) (1) find f (0) (2) prove that f (x) is even function, and f (π - x) = - f (x) (3) if - π / 2 < x < π / 2, f (x) > 0, prove that f (x) monotonically decreases on [0, π]_ ∩)O~

It should be 2F ((x1 + x2) / 2) f ((x1-x2) / 2) after the equation. It's wrong for you to think so

For any x, y satisfies the following conditions: F (x + y) = f (x) f (y) - f (x) - f (y) + 2 and f (1) = 3, when x > 0
F (x) > 2, denote g (x) = f (x) - 1 1. Prove; G (x + y) = g (x) g (y) 2. If G (x) is not equal to 0 for X belonging to R, prove g (x) > 0, and prove that G (x) is an increasing function. 3. Denote an = f (n), prove an + 1 (1 added in the lower right corner) = 2an-1, and find the general term formula of sequence {an}

(1) G (x + y) = f (x + y) - 1 = f (x) * f (y) - f (x) - f (y) + 1 = f (x) * [f (y) - 1)] - [f (y) - 1] = [f (x) - 1] [f (y) - 1] = g (x) * g (y) (2) let x = 0, y = 1, and f (1) = f (0) * f (1) - f (0) + 2, that is, 3 = 3f (0) - 3-F (0) + 2, so f (0) = 2 when x > 0, because f (x) > 2, then G (x) = f (x) - 1 > 2-1

On the functional representation y = f (x + 1), the domain of definition is [1,2]. Find the domain of definition of F (x), f (x-3) because 1 ≤
The domain of y = f (x + 1) is [1,2]. Find the domain of F (x), f (x-3)
Because 1 ≤ x + 1 ≤ 2, 0 ≤ x ≤ 1, the domain of F (x) is [0,1]
Because 1 ≤ x + 1 ≤ 2, so - 3 ≤ x-3 ≤ - 2, so the domain of F (x) is [- 3, - 2]

No
The domain is the scope of X
So the domain of F (x + 1) is [1,2]
Then 1

How many groups are there? Please list the calculation method and the final result!

C(10,30)*A(10,10)=[(30*29*28*27*26*25*24*23*22*21)/10!]*10!=30!/20!

Friction formula
A thing weighs 500 cattle. Use 40 cattle to make it move at a constant speed. If you use 30 cattle to pull it, what is the friction force

First of all, you understand that the maximum friction is 40n
So if you pull with 30n, if the object doesn't move, you can't use the dynamic friction formula;
Using the equilibrium condition, friction equals tension,
Conclusion: friction = 30n

There are two rules to find in mathematics of grade five in primary school: 2,3,4,6,6,9,8, (), () 1,1,2,2,4,4,7, (), ()
Two rules for finding mathematics in grade five of primary school: the process of solving problems (thinking)

2,3,4,6,6,9,8,（12）,（10）
The odd items are: 2, 4, 6, 8, 10 Continuous even numbers
Even numbers are: 3, 6, 9, 12 Continuous multiples of 3
1,1,2,2,4,4,7,7,（11）,(11 )
A group of two numbers is the same number
The number difference of adjacent groups is: 1, 2, 3, 4
For example: 2-1 = 1,4-2 = 2,7-4 = 3

First question: three continuous natural numbers, the middle one is a, what are the other two numbers?

A-1,A,A+1

Given a = [√ (b-2) + 4 √ (2-B)] / (B + 3) + B & # 179 / 2, find the cubic root of ab
It's a process

∵a＝［√（b－2）＋4√（2－b）］/（b＋3）＋（1/2）b^3,
The following results are obtained: a = 0 + (1 / 2) × 2 ^ 3, ab = 2 ^ 3,
∴（ab）^（1/3）＝2.

Simple calculation of 169 + 68 + 32

169+68+32
=169+(68+32)
=169+100
=269

(x to the second power - 2x-1) * (- 2XY) reduction

(x to the second power - 2x-1) * (- 2XY)
=-2x³y+4x²y+2xy