Please help to prove the dual law of sets, A. B. C is any three sets, please help to prove the duality Law: (a ∩ b) ^ C = a ^ C ∪ B ^ C Second, I understand your theory. How can I prove the following question? Let F: X → y, set a belong to set X, and set B also belong to X f(A∪B)=f(A)∪f(B) Thanks for quwozx's answers to these two questions. I understand the first one, but I still don't understand the second one. Can you tell me more about it,

Please help to prove the dual law of sets, A. B. C is any three sets, please help to prove the duality Law: (a ∩ b) ^ C = a ^ C ∪ B ^ C Second, I understand your theory. How can I prove the following question? Let F: X → y, set a belong to set X, and set B also belong to X f(A∪B)=f(A)∪f(B) Thanks for quwozx's answers to these two questions. I understand the first one, but I still don't understand the second one. Can you tell me more about it,

Wen's diagram can be used to help analyze the meaning of the problem and clarify the idea; but it is not rigorous to regard it as the proof process. Next, I give the algebraic proof process. Proof: a ∩ B < AA ∩ B < B ∩ B (a ∩ b) ^ C > A ^ C (a ∩ b) ^ C > b ^ C ∩ B ^ C > A ^ C ∪ B ^ C In the same way, (a ∪ b) ^ C < A ^ C

Proof of set duality law
x∈(A∪B)^c
→ x ∈ a ∪ B
→ X does not ∈ A and X does not ∈ B
→ x ∈ a ^ C and X ∈ B ^ C
→x∈A^c∩B^c
Why is the conclusion of the above proof:
(a ∪ b) ^ C is contained in a ^ C ∩ B ^ C instead of
(A∪B)^c =A^c∩B^c?

In order to prove that two sets AB are equal, it is generally necessary to prove that a contains B and B contains a
So a = B
It is proved that (a ∪ b) ^ C is included in a ^ C ∩ B ^ C, and then it can be explained after proving that a ^ C ∩ B ^ C is included in (a ∪ b) ^ C
(A∪B)^c =A^c∩B^c

Seeking factorization of junior high school
With unknowns
A few more, not from the textbook

After reading the textbook well, you can understand it. Do more exercises. After finishing, summarize your ideas. Do it and do it
Work harder. Mathematics is improving very fast, unlike liberal arts, which can only be seen after a long time

3x square - 6x + 2

Original formula = 3x & sup2; - 6x + 3-1
=(√3x-√3)²-1
=(√3x-√3+1)(√3x-√3-1)

1 (X-1)(x+2)(X-3)(X+4)+24
2 x + xy-6y + X + 13y-6

1. The original formula = [(x-1) (x + 2)] [(x-3) (x + 4)] + 24
=(x^2+x-2)(x^2+x-12)+24
Let x ^ 2 + X-2 be a
(x^2+x-2)(x^2+x-12)+24=a(a-10)+24=a^2-10a+24=(a-4)(a-6)
Replace a
The original formula = (x ^ 2 + X-6) (x ^ 2 + X-8) = (x + 3) (X-2) (x ^ 2 + X-8)
(substitution method)
2. The original formula = (x + 3Y) (x-2y) + X + 13y-6
=(x+3y-2)(x-2y+3)
(double cross multiplication)
Note: x ^ 2 is the square of X, and so on

6x4+7x3-36x2-7x+6 4x4-4x3+13x2-6x+9

Question 1: original formula = 6 (x4 + 1) + 7x (x2-1) - 36x2
=6〔(x4-2x2+1)+2x2〕+7x(x2-1)-36x2
=6[(x2-1)2+2x2]+7x(x2-1)-36x2
=6(x2-1)2+7x(x2-1)-24x2
=[2(x2-1)-3x〕〔3(x2-1)+8x]
=(2x2-3x-2)(3x2+8x-3)
=(2x+1)(x-2)(3x-1)(x+3)．

Extracurricular questions 25 * (X-Y) * (X-Y) + 10 * (Y-X) + 1
If you can, please tell me the classical formula of factorization in Grade 8

Square difference formula: A & sup2; - B & sup2; = (a + b) (a-b)
Complete square formula: A & sup2; ± 2Ab + B & sup2; = (a ± b) & sup2;
There is no specific formula for cross multiplication
In the above question, replace 25 * (X-Y) * (X-Y) with [5 (X-Y)] & sup2;, + 10 * (Y-X) becomes - 10 * (X-Y)
The original formula = [5 (X-Y)] & sup2; - 2 * 5 (X-Y) + 1 & sup2; = [5 (X-Y) - 1] & sup2;

Ask for advice
1: Xsquare-5x-3
2: 3x square + X-2

1: X ^ 2-5x-3 can't be decomposed into the product of two rational coefficients. We can only find out the two of x ^ 2-5x-3 = 0 and write out the result
2：3x^2+x-2=3x^2+3x-2x-2=3x(x+1)-2(x+1)=(3x-2)(x+1)

Square of 1 x-4xy + square of 4y-x + 2y-6
2 ab (square of c-square of D) - CD (square of A-Square of B)

1 x²-4xy+4y²-x+2y-6
=(x-2y)²-(x-2y)-6
=(x-2y-3)(x-2y+2)
2 ab(c²-d²）-cd(a²-b²）
=abc²-abd²-a²cd+b²cd
=(abc²-a²cd)+(b²cd-abd²)
=ac(bc-ad)+bd(bc-ad)
=(bc-ad)(ac+bd)

Two factorizations,
1：9-x^2-6x+9y^2
2：x^2+8xy+16y^2+3x+3y+2

1.(3y+x+3)(3y-x-3)+18
2.(x+4y)²+3(x+y)+2