High School Math Set Topics Partnership Set

{-1},{1},{1 3, 3},{1 2, 2} are either all or none of the elements in the four subsets, so there are a total of 2 4 = 16 and 15 non-empty.

Sets, or sets for short, are a fundamental concept in mathematics and set theory.

The main object of study. The basic theory of set theory was created in the 19th century, and the simplest way to say about a set is the definition in naïve set theory (the most primitive set theory), that is, a set is "a definite set of things", and the "things" in a set are called elements. Modern collections are generally defined as:

A whole made up of one or more definite elements.

Characteristic. Certainty.

Given a set, any element that either belongs to the set or does not belong to the set must be one or the other, and no ambiguity is allowed.

Heterogeneity. Any two elements in a set are considered to be different, i.e. each element can only appear once. Sometimes you need to characterize the situation where the same element appears more than once, you can use a multiset where the element is allowed to appear more than once.

Disorder. In a set, the status of each element is the same, and the elements are disordered with each other. Ordinal relationships can be defined on the set, and after the ordinal relationships are defined, the elements can be sorted according to the ordinal relationships.

But as far as the nature of the set itself is concerned, there is no necessary order between the elements.

Solution: There should be 15 answers.

Method 1: Enumeration.

Contains only one element: {-1}, {1}

There are only two elements: {-1,1},{1 2,2},{1 3,3} contains only three elements: {-1,1 2,2},{1,1 3,3},{1,1 2,2},{1,1 3,3}

It contains only four elements: {-1,1,1 2,2},{1,1,1 3,3}},1,1,1 2,2}, and only contains six elements: {-1,1,1 2,2,1 3,3} Method 2: {-1},{1},{1 3, 3},{1 2, 2} These four subsets of elements are either all or non-take, so there are a total of 2 4=16 possibilities, excluding the empty set, there are 16-1=15 non-empty elements.

Elements 1, -1, 3, 2, 1 2, 1 3 all meet the requirements, choose B

1 and 1 are both about themselves and are counted twice 1 3 and 3, 1 2 and 2 are reciprocal to each other Count twice for a total of four times Because it is a non-empty set, there is 1c4+2c4+3c4+4c4=15?Miscalculated?

Set a=, set b=

If a=b is discussed below:

1) If a=b, then a2=b2

As long as a≠1,a 2≠1,a≠a 2 are satisfied

i.e. a≠-1,1,0

So 1+a 2+b 2=1+2a 2 1 and 1+a 2+b 2=1+2a 2≠3

i.e. the answer can be any number in the set.

2) If a=b 2, then a 2=b

So b 4 = b

Therefore, b=0 or b=1 (there are two complex solutions, if you want, you can write to you, I only write the real number solution).

b=0, b=, does not satisfy the heterogeneity of the set, and is rounded.

b=1, when b=, does not satisfy the heterogeneity of the set, rounds off.

Let's write out the two plural numbers now.

b = -1 2 + ( 3 2)*i or b = -1 2 - ( 3 2)*i

b=-1 2+( 3 2)*i at a=b=

So 1+a 2+b 2=1+b+b 2=1-1 2+( 3 2)*i-1 2-( 3 2)*i=0

b=-1 2+( 3 2)*i at a=b=

So 1+a 2+b 2=1+b+b 2=1-1 2+( 3 2)*i-1 2-( 3 2)*i=0

In summary, 1+a2+b2 can be any number in the set.

If you don't understand, please hi me and have fun studying!

a=b a=b a 2=b 2 or a 2=b, b 2=a When a 2=b, b 2=a, a=b=1 or a=b=o (does not conform to the mutual anisogeneity of the elements, so it is excluded).

When a=b,a2=b2,it seems like a lot of answers (exclude a=+1 , 1,0).

The question is wrong or difficult...

Because b=c

So x +ax + a = 1, x + (a-1) x-1 = 5 subtract to get a = -x-5

Substituting the above equation gives x=-1, so that a=-4

Let the number of people who only solve A, B and C be x, y, and z

then 1 2(x+y+z)=y+z; x+x-1+y+z<25;

x<=8;

Then, with the data less than 8, "the number of people who solved problem B is twice the number of people who solved problem C and x>y", we can finally know that there are only 6 people who solve problem B.

This problem is easier to do, using the method of combining inequality groups and equations, there are only 6 people who solve problem B, 2 people who only solve problem C, and 8 people who only solve problem A.

*2*2*2*2=32 species.

2. C is not selected, that is, 3 out of 9 people are selected, and these 9 people choose 3, a total of c(9,3)=84 kinds of selection, of which A and B are not selected, there are c(7,3)=35 kinds, then any one of A and B is selected and C is not selected, and there are 84-35=49 combinations.

According to the meaning of the topic, a 2-5a+1=7

Get a = 6 or -1

Again7 does not belong to B

That is, a+1 is not equal to 7

So a=-1

7 a a 2-5a + 1 = 7 a = 6 or -1

A+1=7 is not true when a=6.

a=-1...

I think it's probably not that you don't know how to do this kind of question, but that you can't understand it.

I'll give you a translation of the title!

The set a is actually a = its elements are the solution of this equation.

The set b is actually b=, and its elements are the solutions to this equation.

Then look below.

a=b≠ , indicating that the solutions of both equations are the same, and that they both have solutions.

Then it's time for junior high school questions.

The range of values of a is discussed.

a=｛x |4x+p<0} closed pei lao={x |x<-p 4} and a are contained in the car lift b

The result is -p 4 less than or equal to -1

So p is greater than or equal to 4

p 4> Kai socks-1 and talk about Sun Bang-p Han talk 4>2

P<4 and P<-8

The value range is p<-8