Solving mathematics is certain, solving mathematics, and the process of solving must be adopted

OK. Let the arbitrary function be f(x) and its defining domain (-m,m) is symmetrical with respect to the origin.

Suppose f(x) = f(x) + g(x) f(x) is an odd function and g(x) is an even function, both of which define the domain (-m, m) symmetry with respect to the origin.

f(x) is an odd function and g(x) is an even function.

f(-x)= -f(x) g(x)=g(-x)

f(-x)=f(-x)+g(-x)= - f(x) +g(x)②

Syntagion, get f(x) = f(x) + g(x).

f(-x)= - f(x) +g(x)②

Solution: f(x)=[f(x) -f(-x)] 2 g(x)=[f（x) +f（-x)] 2

So f(x) can be expressed as the sum of an odd function and an even function.

That is, the functions of a defined domain about the origin symmetry can be expressed as the sum of an odd function and an even function, and the definition domain of the odd function and the even function is the same as that of the function definition domain.

If you want to prove it, then so:

Let the arbitrary function be f(x) and its defining domain (-m,m) is symmetrical with respect to the origin.

f（x) = [ f（x) +f（x) +f（-x) -f（-x) ]/2

f（x)+f（-x)] /2 +[f（x) -f（-x)]/2

Let f(x)=[f(x) -f(-x)] 2 , g(x)=[f（x) +f（-x)] 2

And both define the domain (-m, m) symmetrically with respect to the origin.

f（x)=f(x)+g(x）

It is enough to prove that f(x) is an odd function and g(x) is an even function.

f(x)=[f（x) -f（-x)] 2 , f(-x)=[f(-x) -f(x)] 2=- f(x), i.e. - f(x)=f(-x), and the domain is defined symmetrically with respect to the origin.

f(x) is an odd function.

Similarly, g(x) is an even function.

A function that defines the symmetry of the domain with respect to the origin can all be expressed as the sum of an odd and even function.

Case 1: This function is non-odd and non-even.

Non-odd and non-even functions can always be expressed as the sum of odd and even functions. Case 2: Other functions, no.

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No, it depends on the situation.

Solution: Set up team B to green area x every day

400/x - 400/2x =4

i.e. 4x=200

Solution x=50

So 2x=100

Answer: A has a green area of 100 square meters per day, and B has a green area of 50 square meters per day (2) Set up team A to work for x days.

You get x 10

A: A should be arranged to work for at least 10 days.

(1) Team B takes 4 days more than Team A, Team A's efficiency is 2 times that of Team B, the number of days Team B does is 2 times that of Team A, 400 square meters, Team B does 4x2 8 days, and does 400 8 50 (square meters) every day Team A does 50x2 100 (square meters) every day (2).

Linking BE, CE, it is easy to obtain AEF is all equal to AEG, and it is easy to obtain AF=AC+CG, that is, BF=AB-AF=AB-(AC+CG)=AB-AC-CG, and BF+CG=6

It is easy to obtain feb which is equal to gec, and bf=cg=3

The first three acute angles, the last four obtuse angles. Unless c = 10 or root number 28, it cannot be a right triangle.

Solution: 1, (1)64=8 8=2 6 plus or minus 2, (2) 243=81 3=3 5 -3 (3) 0

2, (1) When n is an even number, this number must be positive, and it has 2 roots to the nth power, and they are opposite to each other.

2) When n is an odd number, this number can be either positive or negative, and the nth root of a positive number is a positive number, a negative number.

The nth power root is a negative number.

3) The nth root of 0 is 0

(1).2 -2 (2 6 = 64) The even power of a negative number is equal to the even power of the opposite number.

0(2).The even power roots of positive numbers are opposite to each other.

11，70'

12, three. 13. Tangent.

14, the question is not clear, you can re-shoot it for me.

15, 3 or 4

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