I want to find a set of math handouts for independent enrollment in high schools.

First, knowledge induction:

1) Set: set some specified objects together to form a set. Each object is called an element.

Note: ① Set and its elements are two different concepts, which are given by description in textbooks, similar to the concepts of points and lines in plane geometry.

② The elements in the set are deterministic (A? A and a? A, the two must be one), different from each other (if a? A, b? A, then a≠b) and disorder ({a, b} and {b, a} represent the same set).

③ A set has two meanings: all eligible objects are its elements; As long as it is an element, you must sign the condition.

2) Representation methods of sets: enumeration method, description method and graphic method are commonly used.

3) Classification of sets: finite set, infinite set and empty set.

4) Common number set: n, z, q, r, N*

2. Concepts such as subset, intersection, union, complement, empty set and complete set.

1) subset: if there is x∈B for x∈A, then A B (or ab);

2) proper subset: A B has x0∈B but x0 A；; ; Marked as b (or, and)

3) Intersection: A∩B={x| x∈A and x∈B}

4) and: A∪B={x| x∈A or x∈B}

5) Complement set: cua = {x | x but x ∈ u}

Note: ①? A, if A≦? And then what? a；

(2) If, then;

③ If and, then A=B (equal set)

3. Make clear the relationship between set and element, set and set, master relevant terms and symbols, and pay special attention to the following symbols: (1) and? Difference; (2) the difference between and; (3) The difference between and.

4. Several equivalence relations about subsets

①A∩B = A A B； ②A∪B = B A B； ③A B C uA C uB；

④A∩CuB = empty set cuab5cua ∪ b = iab.

5. The essence of intersection and union operations

①A∩A=A，A∩？ = ? ，A∩B = B∩A； ②A∪A=A，A∨？ =A，A∪B = B∪A；

③Cu(A∪B)= CuA∪CuB，Cu(A∪B)= CuA∪CuB；

6. Number of finite subsets: If the number of elements in set A is n, then A has 2n subsets, 2n- 1 nonempty subset and 2n-2 nonempty proper subset.

2. Examples:

Example 1 given set m = {x | x = m+, m ∈ z}, n = {x | x =, n ∈ z}, p = {x | x =, p ∈ z}, then m, n and p satisfy the relationship.

A) M=N P B) M N=P C) M N P D) N P M

Analysis 1: Start with judging the uniqueness and difference of elements.

Answer 1: For the set m: {x | x =, m ∈ z}; For the set n: {x | x =, n ∈ z}

For the set p: {x | x =, p ∈ z}, because 3(n- 1)+ 1 and 3p+ 1 both represent numbers divided by 3, and 6m+ 1 represents numbers divided by 6.

Analysis 2: Simply enumerate the elements in the collection.

Answer 2: M={…, …}, N={…, …}, P={…, …}. At this time, don't rush to judge the relationship between the three sets, but analyze the different elements in each set.

= ∈N, ∈N, ∴M N, and = M, ∴M N,

= P, ∴N P and ∈N, ∴P N, so P=N, so choose B.

Comments: Because the second way of thinking only stays in the initial inductive hypothesis and does not solve the problem theoretically, I advocate the first way of thinking, but the second way of thinking is easy to handle.

Variant: Set, and then (b)

A.M=N B.M N C.N M D。

Solution:

When 2k+ 1 is an odd number and k+2 is an integer. Choose B.

Example 2 defines the sets A*B={x|x∈A and x B}. If a = {1, 3,5,7} and b = {2 2,3,5}, then the number of subsets of A*B is

1 B)2 C)3 D)4

Analysis: To determine the number of subsets of the set A*B, we must first determine the number of elements, and then use the formula: the set A={a 1, a2, …, an} has 2n subsets to ask for the solution.

Answer: ∫a* b = {x | x∈a and x B}, ∴A*B={ 1,7} has two elements, so A*B * * has 22 subsets. Choose D.

Variant 1: the known nonempty set M {1, 2, 3, 4, 5}, if a∈M, then 6? A∈M, then the number m of sets is

A) Five b) Six c) Seven d) Eight.

Variant 2: Given {a, b} A {a, b, c, d, e}, find the set a. 。

Solution: A known set must contain elements A and B.

Set a can be {a, b}, {a, b, c}, {a, b, d}, {a, b, e}, {a, b, c, d}, {a, b, c, e}, {a, b, d, e}.

Comment on the number of set A in this question is actually the number of proper subset of set {c, d, e}, so * * * has one.

Example 3 sets a = {x | x2+px+q = 0} and b = {x | x2? 4x+r=0}, and A∩B={ 1}, A∪B={? 2, 1, 3}, the values of real numbers p, q, r q, r.

Answer: ∫ a ∩ b = {1} ∴1∈ b ∴12? 4× 1+r=0，r=3。

∴B={x|x2？ 4x+r=0}={ 1，3}，∫A∪B = {？ 2, 1,3},? 2 B，∴？ 2∈A

∫ A ∩ B = {1} ∴/kloc-0 ∈ A ∴ Equation x2+px+q=0 is -2 and1.

∴ ∴

Variant: known sets A = {x | x2+bx+c = 0}, b = {x | x2+MX+6 = 0}, and values of A∩B={2}, A∪B=B, and real numbers B, C and M. 。

Solution: ∫a∩b = {2} ∴ 1∈b ∴22+m? 2+6=0，m=-5

∴b={x|x2-5x+6=0}={2,3}∫a∪b = b∴

∫a∩b = { 2 }∴a = { 2 }∴b =-(2+2)= 4，C = 2× 2 = 4。

∴b=-4,c=4,m=-5

Example 4 the known set a = {x | (x-1) (x+1) (x+2) >; 0}, set b satisfies: a ∪ b = {x | x > -2}, and A∩B={x| 1.

Analysis: simplify the set a first, and then use A∪B and A∪B to determine which elements on the number axis belong to B and which elements do not belong to B.

Answer: A={x|-2 1}. From A∩B={x| 1-2}, we can know [- 1, 1] B, while (-∞,-2) ∩ b = Ф.

Taken together, the above categories are B={x|- 1≤x≤5}

variant 1:If A = { x | x3+2 x2-8x & gt； 0}，B={x|x2+ax+b≤0}，A∪B = { x | x & gt； -4}, a ∩ b = φ, find A and B. (Answer: a=-2, b=0)

Comments: When solving a kind of set problem about inequality solution set, we should pay attention to using the method of combining numbers and shapes to do the number axis to solve it.

Variant 2: Let m = {x | x2-2x-3 = 0} and n = {x | ax- 1 = 0}. If M∩N=N, find all sets that meet the conditions.

Answer: m = {- 1, 3}, ∫m∩n = n, ∴ n m.

① When ax- 1=0 has no solution, ∴a=0 ②.

Synthesis ① ②: The required set is {- 1, 0,}

Example 5 Given a set, the domain of the function y=log2(ax2-2x+2) is q, and if p ∩ q ≠ φ, the value range of the real number A..

Analysis: Firstly, the original problem is transformed into the inequality AX2-2x+2 >; 0, and then solve it by parameter separation.

Answer: (1) If yes, there is a solution in it.

When the time is right,

So a & gt-4, so the range of a is

Variant: If the equation about X has a real root, find the range of the number A. ..

Answer:

Comments: Using parameters to solve problems generally requires classified discussion, but not all problems should be discussed. How to avoid discussion is the key for us to think about this kind of problem.

Three. class exercise

Multiple choice

1. The following eight relationships ①{0}= ② =0 ③ {} ④ {} ⑤{0}

⑥ 0 ⑥ {0} ⑧ {} The correct number.

(A)4 (B)5 (C)6 (D)7

2. proper subset with the set {1, 2,3} has * * *.

Five (b) six (c) seven (d) eight

3. the set A={x} B={} C={} and then there is.

(a) (a+b) A (b) (a+b) B (c) (a+b) C (d) (a+b) Any one of A, B and C.

4. Let A and B be two subsets of the complete set U, then the following formula holds.

(A)CUA cub (B)CUA cub =U

(C)A = (D)CUA B=

5. If the sets A={} and B={} are known, then A=

(A)R (B){ }

(C){ } (D){ }

6. The following statements: (1)0 and {0} represent the same set; (2) The set consisting of 1, 2,3 can be expressed as

{1, 2,3} or {3,2,1}; (3) The set of all solutions of equation (x- 1)2(x-2)2=0 can be expressed as {1, 1, 2}; (4) Set {0} is a finite set, which is correct.

(a) only (1) and (4) (B) only (2) and (3).

7. Let S and T be two nonempty sets, and S T, t S, let X=S, then S∪X=

X(B)T(C)φ(D)S

8 Let the unary quadratic equation ax2+bx+c = 0 (a

(A)R (B) (C){ } (D){ }

fill-in-the-blank question

9. In rectangular coordinate system, the point set on the coordinate axis can be expressed as

10. if A={ 1, 4, x}, B={ 1, x2} and A B=B, then x=

1 1. If A={x} B={x} and the complete set U=R, then A=

12. If the equation 8x2+(k+ 1)x+k-7=0 has two negative roots, then the value range of k is

13 let the sets A={}, B={x}, A B, then the value range of the real number k is.

14. let the complete set U={x is a nonnegative odd number less than 20}, if a (cub) = {3 3,7, 15}, (cua) b = {13, 17,/kloc-.