A math problem about a set problem.

Categorize 1 to 50 and divide them into 7 divided by 7 and divisible by 7, with 8 remaining 1 and 1 remaining and 7 others. In the same way, the elements of the remaining 2 and the remaining 5 cannot exist at the same time, the remaining 3 and the remaining 4 cannot exist at the same time, and the divisible can only exist in one element at most, so at most there are 8 remaining 1, the remaining 2 or 5 choose one category, the remaining 3 or 4 choose one category, and the divisible one can be selected, a total of 23.

Dividing 1 to 50 can be divided by 7 and divisible by 7, with 8 remaining 1 and 7 others.

From the meaning of the title: the elements of the remaining 1 and the elements of the remaining 6 cannot exist at the same time, in the same way, the remaining 2 and the remaining 5 cannot exist at the same time, the remaining 3 and the remaining 4 cannot exist at the same time, and the divisible can only exist in one element at most, so the maximum is 8 of the remaining 1, the remaining 2 or 5 choose one category, the remaining 3 or 4 choose one category, and the divisible one is selected, a total of 8 + 7 + 7 + 1 = 23.

Put 1, 2, 3 ,......50 per 7 number group can be divided into:

1 6 , 2 5 , 3 4 , 8 13 , 9 12 10 11 ,...... can be found43 48, 44 47, 45 46 are all divisible by 7, so these 21 logs can only be all in the set s.

Take the previous number or both take the next number, i.e. (1,2,3

8,9,10……43, 44, 45) or (4, 5, 6, 11, 12, 13....46,47,48), in the set s there can only be multiples of 7.

One, plus the number 50, is the most in the set s.

I can have 23 numbers.

Regarding the first question, since we are talking about a set, and the x in the set must satisfy the square of x plus 2 to equals 0, such an x does not exist, so such a set is an empty set. It's just an empty set, and it can't be said to be meaningless.

On the second question, the main difference between the two sets is the parentheses. In the first case, the numbers are covered with small parentheses, which means that the two numbers inside are the abscissa and ordinate of a certain point, so the first set element is the point (1,2), and the second set element is the point (2,1), which is naturally different. In the second case, there are no parentheses, indicating that the numbers in the set are all his elements, the first set has 4,5, and the second set has 5,4.

The two set elements are the same, so the set is naturally the same, regardless of the order of the elements.

x squared + 2 is the constant of 2, and it doesn't make sense to take any number.

The first sentence is true, x 2+2=0, this equation has no solution in the range of real numbers, because the square of any real number cannot be equal to -2, that is, it is meaningless in the range of real numbers.

The second sentence is wrong: notice that both sets are sets of points, and the points in the set are (1,2) and the other is (2,1) are not the same, and the elements are not the same for the set, so they do not represent a set.

The third sentence is true, both sets are numbers, 4, 5 and 5, 4 are no difference, so yes, I am a high school math teacher, can be in contact for a long time, if you have any questions, please direct me, I hope your grades improve.

In the range of real numbers is an empty set.

With are two different points and are different sets.

The same collection as represented.

The x-squared must be greater than 0. If x square 2 0, x square = i times the root number 2, this is an imaginary number.

It is the same set as the representation, for example, eating the main food first and eating the vegetables first are both called eating.

The same collection as the expression, such as putting water first and tea leaves first, is called making tea.

Idea analysis: The meaning of the question is intuitively represented by a venn diagram, the conditions are intuitively expressed, and the problem is easily solved.

15-3-3 = 9 (people) who only participated in swimming competitions;

Those who only participate in track and field competitions have 8-3-x=5-x (people);

Those who only participate in ball games have 14-3-x = 11-x (people);

There are 3 people, 3 people, and X people who participated in the two competitions at the same time;

Therefore, 9+(5-x)+(11-x)+3+3+x=28, and the solution is x=3

Therefore, there are 3 people who participate in both track and field and ball games, and 9 people who only participate in swimming competition.

15 + (14-3) + (8-3) - 28 = 3 people So there are 3 people who participate in track and field and ball games at the same time.

15-3-3 = 9 people So there are only 9 people who participate in swimming.

This kind of inscription can be made with a Wayne diagram.

The easiest way to do this is to draw a picture.

If you make 4 circles, which represent the total participation, swimming, track and field, and ball, then the intersection of swimming and track and field is 3, the intersection of swimming and ball is 3, and the intersection of swimming, track and field, and ball is 0Then there is the following equation.

15 + 8 + 14-28 = 9 people, 9-3-3 = 3 people, this is the number of people participating in athletics and ball games.

28-8-14+3=9 people. This is the number of people who participate in swimming only.

Choose A, the process hi me, I'll tell you.

x and y are both odd sets, x+y is an even number, m is an even set, so choose a

b is an option, m is an even set, x is an odd set, and y is a subset of an odd set, which is when k is an even number.

x is all odd y odd then x+y is an even number m represents all even numbers, so choose a

The number brought in can also be proved that if k=1 then, x=3, y=5 then x+y=8, then we can see that 8 is an even number, so m

Choosing b because x is seen as a set of odd numbers, and y is seen as a set of fours divided by 1 (which is certainly also part of an odd number).

Then y is a subset of x, additive, or x

x -5x+9=3, solve x

x + ax + a = 2 and x + ax + a = 4 to solve a and x;

x + ax + a = 1 and x + (a + 1) x-3 = 3 to solve a and x specific steps to solve by themselves, the method refers to the solution of the equation.

(1) If the set is equal, then the elements in the set must be equal. Then there is x -5x+9=3 and the solution is x=2 or x=3

2) 2 belongs to b, then x +ax + a = 2

and b is a true subset of a, then 3 belongs to a, and there can only be x + (a + 1) x-3 = 3, according to these two equations to solve a , x

3) Obviously, 1≠3, then only x +ax + a=1 and x + (a+1)x-3=3 can solve a and x

n(m) indicates the number of elements in the set m.

n(m)=3, that is, the number of elements is 3.

x-5)(p-x)>0;i.e. (x-5)(x-p)<0;

If p<5; The solution of the inequality is: p5; The solution of the inequality is: 5

This means that the set m is greater than zero and px is an integer.

（x-5）(x-p)<0

5 The first case.

p=9, the second case.

p = 1, so p can only take 1 or 9

If a b = empty set.

then a is a subset of b's complement.

i.e. {x|2a x a+3} is {x|-1 x 5} has 2a -1 and a+3 5

Solution: -1 2 a 2

There is no solution, i.e. the range of a is an empty set.

Draw a number line to find out.

52a>-1

a+3>2a

The solution is -1 2=5

2a<=-1

a+3>2a

There is no solution. a is an empty set.

1 a+3<5

2a>-1

a+3>2a

The solution is -1 2=5

2a<=-1

a+3>2a

There is no solution, so a is an empty set.

b = (- 1) (5, + a) when 2a a+3,===>a 3, it is easy to know that a is empty, satisfying a b = empty. When 2a a+3, there should be -1 2a a+3 5

=>-1/2≤a≤2.Therefore a [-1 2,2] (3,+, a b = empty. (2) There should be 2a -1 5 a + 3, ===> -2 a -1 2

So when a (-2, -1 2), a b = r

1 a b = empty set, then a + 3<5

and 2a>-1

and a+3>2a

The solution is -1 2=5

and 2a>-1

and a+3>2a

There is no solution, so a is an empty set.

From 10-x s, we know that the element in s must be less than 10 (if it is greater than or equal to 10, then 10-x is less than or equal to 0 and does not meet condition a).

Then it is from 1, 2, 3, 4, 5, round noise 6, 7, 8, 9 these 9 remainders of Peitong.

However, from condition B, if you choose 1, you must choose 9

If you choose 2, you must choose 8

Therefore, it is equivalent to choosing any 1 5 out of the 5 numbers 1, 2, 3, 4, and 5 to be used as the element of s.

This selection method is obtained by the number of permutations and combinations = 5 + 10 + 10 + 5 + 1 = 31.

This selection method is obtained by the number of permutations = 5+

A few more elements are both true and subset, and you can figure it out by figuring out the definition and doing more questions.