Given the set a = {1,2,3,4,5}, and the set B = {4,5,6,7,8,9}, we can find a ∩ B and AUB,

Given the set a = {1,2,3,4,5}, and the set B = {4,5,6,7,8,9}, we can find a ∩ B and AUB,

A∩B=｛4,5｝AUB=｛1,2,3,4,5,6,7,8,9｝

Given that sets a and B belong to set {1,2,3,4}, then the condition a ∩ B = {1,2,3,4} is satisfied,

Set a and B belong to set {1,2,3,4}, which is wrong. It should be said that [set a and B are contained in set {1,2,3,4}]. Two sets a and B satisfying condition a ∩ B have numeral elements 1 and 2. Therefore, a = {1,2}. B = {1,2,3}. A = {1,2,3}. B = {1,2}. A = {1,2,4}. B = {1,2}. A = {1,2,4}

We know that set a = {2,4,6,8,9}, B = {1,2,3,4,5,8}, and that set C is such a set: if all elements of set C add 2, it becomes a subset of a; if all elements of set C subtract 2, it becomes a subset of B. try to write such a set, C =________ ．

Subtract 4 from the elements in a, and then find the intersection with B
A - 4 = {-2,0,2,4,5}
A intersection B = {2,4,5}
So C can be {2,4,5} + 2 = {4,6,7}

How about 2006 divided by 2006 and 2006 out of 2007

2006 (2006 and 2006 / 2007)
=2006 / 2007 (2007 × 2006 + 2006)
=2006 / 2007 × (2007 + 1)
=2007 (2006 × 2008)
=2007 / 2008

Given a & sup2; + B & sup2; + C & sup2; - 2A + 4B + 14-6c = 0, find the value of (a + B + C) & sup2

Given a & sup2; + B & sup2; + C & sup2; - 2A + 4B + 14-6c = 0, find the value of (a + B + C) & sup2
a²+b²+c²-2a+4b+14-6c=(a-1)²+(b+2)²+(c-3)²-1-4-9+14=(a-1)²+(b+2)²+(c-3)²=0
So a = 1, B = - 2, C = 3
∴(a+b+c)²=(1-2+3)²=4

2 (2A2 + 9b) - 3 (5a2-4b) where a = - 1, B = 12

The original formula is 4a2 + 18b-15a2 + 12b = - 11A2 + 30b; when a = - 1, B = 12, the original formula is = - 11 × (- 1) 2 + 30 × 12 = 4

Factorization of - 4A & sup2; B & sup2; + 8A & sup2; b-2a
Please give your answer in two days,

－4a²b²＋8a²b－2a
Proposed - 1
=-(4a²b²-8a²b+2a)
Proposed 2A
=-2a(2ab²-4ab+1)

Merge similar items and remove brackets
Try to explain that the value of the algebraic formula - 3 (ab-2a) + 5 (B-4) - (- 3AB + 6a-19) has nothing to do with the value of the letter A

-3（ab-2a）+5(b-4)-(-3ab+6a-19)
=-3ab+6a+5b-20+3ab-6a+19
=-3ab+3ab+6a-6a+5b-20+19
=5b-1
So it has nothing to do with a