Set u = {1,2,3,4,5,6}, B = {1,4}, a = {2,3,5}, find ∁ U (a ∪ b) and (∁ UA) ∩ (∁ UB)

Set u = {1,2,3,4,5,6}, B = {1,4}, a = {2,3,5}, find ∁ U (a ∪ b) and (∁ UA) ∩ (∁ UB)

If ∁ u = {1,2,3,4,5,6}, B = {1,4}, a = {2,3,5}, ∁ UA = {1,4,6}, ∁ UB = {2,3,5,6}, a ∪ B = {1,2,3,4,5}, then ∁ U (a ∪ b) = {6}, (∁ UA) ∩ (∁ UB) = {6}

It is known that a ∩ B = {3}, 9 (CUA) ∩ B = {4,6,8}, a ∩ (cub) = {1,5}, (CUA) U (cub) = {x | X

9 (CUA) ∩ B = {4,6,8}, of which 9 is the most
A ∩ B = {3}, (CUA) ∩ B = {4,6,8}, we get that there is {3,4,6,8} in B
A ∩ B = {3}, a ∩ (cub) = {1,5}
(CuA)U(CuB)={x|x

Cu (A and b) = (CUA) intersection (cub) is analyzed by Wayne diagram
As soon as possible,

As shown in the figure, it is found that the upper and lower figures are the same
So (CUA and b) = (CUA) intersection (cub)

How to prove the dual law of proof set

It is proved that a ∩ B ∩ AA ∩ 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 ∩ B ^ C substitutes a ^ C into a, and B ^ C into B, thus (a ^ C ∪ B ^ C) ^ C ＜ (a ^ C) ^ C ∩ B ^ C = a ∩ B ∩ two sides complement, a ^ C ∪ B ^ C ＞ (a ∩ B ∩ B ∩ B ∩ C ∩ B ∩ C ∩ C ∩ C ∩ B ∩ C ∩ C

(factorization)
xy-x+1-y
Well, I'm a stupid student-_ -111

(x-1)*(y-1)

Factorization problem in grade two (for me)
If 4x & sup2; - 4x + 9y & sup2; - 12Y + 5 = 0, find the value of 6x + 2 / 3Y

(2x-1)^2+(3y-2)^2=0
x=1/2,y=2/3
6x+2/3y=3+1=4

Urgent problem of factorization in grade two~
If one part of x minus one part of Y equals one part of 8, then 2Y + 3xy-2x =?

1/x-1/y=1/8
(y-x)/xy=1/8
xy=8(y-x)
2Y + 3xy-2x of 7y-7x-xy
=26(y-x)/-(y-x)
=-26

A difficult problem related to factorization
It is known that a, B and C are the three sides of a triangle, and satisfy the following conditions: the square of a + the square of B + the square of C = AB + BC + ca

Multiply two sides by two
(a-b)^2+(b-c)^2+(a-c)^2=0
So a = b = C
So it's an equilateral triangle

Division and factorization of integers in grade two
Given X & sup2; - 3x + 1 = 0, find the value of X & sup2; + (1 / X & sup2;)
Given that | ab-2 | and (B-1) & sup2; are opposite numbers, try to find the value of 1 / AB + 1 / (a + 1) (B + 1) + 1 / (a + 2) (B + 2) +. + 1 / (a + 2008) (B + 2008)
I hope both questions can be answered
Try to keep the process fresh

X & sup2; - 3x + 1 = 0 two sides divided by X, we can get x-3 + 1 / x = 0 + 1 / x = 3 two sides square, we can get X & sup2; + 1 / X & sup2; = 9-2 = 72. Because the absolute value of any number is nonnegative, the square is nonnegative, so ab-2 = 0, B-1 = 0, so B = 1, a = 2, so 1 / AB + 1 / (a + 1) (B + 1) + 1 / (a + 2) (B + 2) +. + 1 / (a + 2008) (B + 2008) = 1 / 1 * 2 + 1 /

Several factorizations
(XY)^2+2XY+1-(X+Y)^2
(2X-3Y)^3+(3X-2Y)^3-125(X-Y)^3
(X^4-4X^2+1)(X^4+3X^2+1)+10X^4

(XY)^2+2XY+1-(X+Y)^2=(xy+1)^2-(x+y)^2=(xy+1-x-y)(xy+1+x+y)(X^4-4X^2+1)(X^4+3X^2+1)+10X^4 =(X^4+1)^2-x^2(x^4+1)-2x^4=(x^4+1-2x^2)(X^4+1+x^2)=(x^2-1)^2（x^2+1）^2=(x+1)^2(x-1)^2(x^2+1)^2 (2x-3y)^3+(3x-2...