100,000 urgent high school math set questions

cua)∩b=

x²+（a-3）x+a²-3a=0

Composite as a system of equations.

Get a=2 or a=0

a=0 substitution does not match rounding.

a= b=a∪b=

From the question a=b= so a is a subset of bde.

So a b = a (cua) b = cba

1.Because x is divisible by 3 and n is divisible by 2, x is divisible by 2,3, i.e. x is a multiple of 6;

In the same way, we can know that b is a multiple of 2. Then the intersection of a and b means that they both have something in common, because 6<24 so the set a is included in the set b, so they are divisible by 6 in a and less than 24;

1.Because (n 2) n, then there is 6*(n 2) n, so 3n n, then a;

Since (24 m) n,, then there is m=1,3,8,12,24 then b so there is a b a cua=

cua)∩b＝ø

2.From a≠b and a b≠, we can see that a and b have at least one element identical, i.e., the equation x + (2a-3) x-3a = 0 and x + (a-3) x + a -3a = 0 have at least one identical root.

1) If two equations have only one of the same roots, let one of the same roots be b;

Then there is b + (2a-3) b-3a = b + (a-3) b + a -3a simplified to obtain, b = a

Substituting b=a into the equation x + (2a-3)x-3a=0 to solve a 2, and substituting a 2 into the equations x + (2a-3)x-3a=0 and x +(a-3)x+a -3a=0 respectively to solve a and b

a＝;b＝;

A b can be obtained

2) If the two equations have two identical roots, then there is 2a-3=a-3 and -3a a -3a to obtain a=0, and the two equations are substituted to solve a and b

a＝b＝;So a b

If the five numbers are different, take any 4 numbers, then there are 5 ways to take them.

However, the sum of the set has only 4 elements, which means that 2 out of 5 numbers are the same.

Then there are 4 cases of the sum of these 5 numbers.

44 + 45 + 46 + 47 + 44) 4, do not meet the meaning of the topic and give up.

44 + 45 + 46 + 47 + 45) 4, does not meet the meaning of the topic and leaves.

44 + 45 + 46 + 47 + 46) 4 = 57, which is in line with the title.

44 + 45 + 46 + 47 + 47) 4, which does not meet the meaning of the topic.

Subtract the elements in the set with to get 4 numbers, 13, 12, 11, 10.

and 10+11+12+13=46 is 1 smaller than 47 and 211 larger than 44 The fifth number is 11

It can only be 11, and these 5 numbers are 10, 11, 11, 12, 13

x1, x2, x3, x4, x5 take any four numbers, and the sum of them consists of four numbers.

So x1, x2, x3, x4, x5 have two equal numbers.

44+45+46+47+z) 4 is an integer to fit the meaning of the topic (z=44, 45, 46, 47).

z=46, these five numbers are 10, 11, 11, 12, 13

First of all, the set of the sum obtained by taking any four of the five positive integers has only four elements, and the five positive integers are far more than four for each of the four positive integers, from which it can be inferred that two of the five positive integers are identical, and we can let the sum of the repeated numbers be z, then the following equation holds:

x1+x2+x3+x4+x5) 4 44+45+46+47+z It can be seen that the left side of the equation is a multiple of 4.

So z1 must be a multiple of 4 that is 2 more than that.

For the equation to hold, z can only be equal to 46

So the sum of the five positive integers is 57

The five positive integers are 10, 11, 11, 12, and 13

A total of five sets can be formed, and they can be discussed separately. For example, these elements can be x1, x2, x3, x4,.

A set is actually a set of real numbers, except that it has only one real number 1, so it is equivalent to a set.

Not an equal set.

The first set is the number x, and the value of x is 1

The second set contains an equation of x=1, which is not the same.

Collections can contain numbers, algebraic formulas, and even equations.

So be careful.

And is the inclusion relationship, containing. Because x can be squared by 1 or -i (expressed as a complex number).

The first few "weird numbers" need to be tried, and it should be noted that 1 is neither prime nor composite. It is easy to see that 9, 11, and 13 are all "weird numbers". In fact.

For 9:2,3,5,7 are prime numbers; 4, 6, 8, 9 are composite numbers.

For 11: 2, 3, 5, 7, 11 are prime numbers; 4, 6, 8, 9, 10 are composite numbers.

For 13: 2, 3, 5, 7, 11, 13 are prime numbers; 4, 6, 8, 9, 10, 12 are composite numbers.

Let's illustrate that there are only these three "weird numbers", i.e., a=

Counting down from 13 is 14, 15, 16, and all three are composite. So by 16, there are three more composite numbers than prime numbers. Every time a prime number is counted in the future, its next number is an even number, and it must be a composite number.

Therefore, starting from 17, for every positive integer n, the number of composite numbers is at least two more than the prime number of integers not exceeding n, i.e., it is impossible to be equal. So 9, 11, and 13 are all "weird numbers".

Oh yes, 1 yes too, I ignored it. Because a positive integer that does not exceed 1 has only 1 itself, and the number of prime and composite numbers is 0, 1 is also a "weird number".

Solution: a===

Because a b=

When m+1 2m-1, i.e., b= time, m2 when m+1 2m-1, i.e., m2, 2m-1 -2 or m+1 5 solution gives m -1 2 or m 4

And because m 2, m 4

In summary, the value range of m is m 2 or m 4

This needs to be drawn to get the inequality.

Let the number of participants in the track race be A, the set of the field race as B, and the ball game set as C, the number of people who only participate in A is A, the number of people who only participate in B is B, the number of people who only participate in C is C, and then the number of people who participate in both A and C is X, and then the following equation can be drawn from the title:

a+2+x=15

2+1+2+b=8

c+x+2=14

a+b+c+4+x=27

Combining the above equations, we can solve:

a=8b=3

c=7x=5

Therefore, the number of people participating in both track and ball races is 5, and the number of people participating in only one race is a+b+c=18.

According to the diagram, it can be seen that the number of people who only participate in field is 8-2-2-1=3 people, and if only X people participate in the track category, only Y people participate in the ball game, and the number of people who participate in the ball game and the track class is Z people, then x+z+1+2=15

y+z+1+2=14 ②

x+y+z=27-8 ③

From the above equation we get x=8 y=7 z=4

That is, only 8 people participated in the track category, 7 people only participated in the ball game, 8-2-2-1 = 3 people only participated in the field category (as drawn from the figure), and 4 people participated in both the ball game and the track category.

I drew a simple diagram.

A means only to participate in field races, B only to participate in track races, C only to participate in ball games, and X means to participate in track races and ball games.

It is known that the total number of people is 27, a+b+c+2+2+1=27a+2+2+1=8

b+2+1+x=15

c+2+1+x=14

Four equations and four unknowns can be solved.

Answer: a=3, b=8, c=7, x=4

Solution: People who participate in track and ball games at the same time = 15-3 = 12 people who participate in both track and field games and ball games = 8-3 = 5

If there are x people who participate in both track and ball races and do not participate in track and field races, then the number of people who only participate in track races = 12-x

People who only participate in the ball game = 14-3-x = 11-x

Yes: 27 = 12 + 11 - x + 8

x=4 then:

Number of people who only participate in the track race = 8

People who only participate in the ball game = 7

The number of people who only participate in the field competition = 2

So people who participate in only one competition = 8 + 7 + 2 = 17

Suppose there are x people who participated in both the track race and the ball game, and the equation 15-x-3+8+14-x-3+x=27 can be solved, and x=4 can be solved.

The left side can be changed as: ((root number 3)-1) (root number 2) - (root number 3) + 1) (root number 2) = root number 6

So it belongs to the set (a=0 and b=1).

The key is that the numerator and denominator in the root number are multiplied by 2, and then the numerators are combined to form a perfect square with the root number.

Solution: t= (2- 3)+ 2+ 3)Then t = (2- 3) + (2 + 3) + 2 [(2- 3) (2 + 3)] = 6

=>t=√6=0+1×√6.(a=0,b=,b∈q).Therefore, it can be known from the inscription.

Fill in belonging).

Let (2- 3)+ 2+ 3)=x

2- 3)+ 2+ 3)= The squared of x yields (2- 3)+ 2+ 3)= 6Answer: Contains.

In the first case, b+a=ac and a+2b=ac 2, so c=(a+b) a is substituted into 2.

a^2+2ab=（a+b）^2

So b 2=0 b = 0 but a+b cannot be equal to a so the first case is not true.

In the second case, b+a=ac 2 and a+2b=ac, so c=(a+2b) a is substituted into 2.

a^2+ab=（a+2b）^2

b(4b+3a)=0 Because b is not equal to 0, so 4b=-3a, so c=(a+2b) a=(

See that the numbers in b are proportional.

If b+a corresponds to AC and a+2b corresponds to AC2, then (a+b) 2=a(a+2b)--b=0, contradicts the heterogeneity of the set and is rounded.

So only B+A corresponds to AC 2, and A+2B corresponds to AC, then (A+2B) 2=A(A+B)--B=-3A 4

So a+2b=-a 2---a+2b corresponds to ac] --c=-1 2

This set problem takes advantage of the mutual heterogeneity of the elements, the concept of the equality of sets. A=B requires B+A=AC and A+2B=AC

Or b+a=ac and a+2b=ac, the first one finds c=1, b=0, and both a and b are not mutually heterogeneous, so they are rounded.

The second way to find the relation of ab is to bring in a+2b=ac to find c, and that's it.

From a=b, ac=a+b and ac2=a+2b, we get c=1 (which is not the case, because when c=1, the three elements in set b are the same).

So only when ac=a+2b and ac=a+b, the solution is c=-1 2