Let a = {x | X be a positive integer less than 9}, B = {1,2,3}, C = {3,4,5,6}, find anb, ANC, an (BUC), Au (BNC,

Let a = {x | X be a positive integer less than 9}, B = {1,2,3}, C = {3,4,5,6}, find anb, ANC, an (BUC), Au (BNC,

1. A intersection B = {1,2,3}
2. A intersection C = {3,4,5,6}
3. A intersection (B and C) = {1,2,3,4,5,6}
4. A-union (b-intersection C) = {1,2,3,4,5,6,7,8}

Prove an (B-C) = (anb) - (ANC)

We should eliminate A1,. An, leave 2 ^ 2,. 2 ^ n, etc
Tn=2an＋2 a(n-1)＋2∧3a(n-2)＋… ＋2∧na1①
2Tn=2 an＋2∧3a(n-1)＋2∧4a(n-2)＋… 2∧(n＋1)a1②
The dislocation subtraction method an = 3n-1, that is an-a (n-1) = a (n-1) - A (n-2) =. = a2-a1 = 3
②-①Tn=-2an＋2 [an-a(n-1)]＋2∧3[a(n-1)-a(n-2)]＋… ＋2∧n[a2-a1]+2∧(n＋1)a1
=-2an＋3*2 ＋3*2∧3＋… ＋3*2∧n+2∧(n＋1)*2
=-2an＋3(2 ＋2∧3＋… ＋2∧n)+2∧(n＋2)
=-2an+3[-4+2^(n+1)]+2∧(n＋2)
=-2(3n-1)-12+3*2^(n+1)+2*2∧(n＋1)
=5*2^(n+1)-6n+2-12
Left = TN + 12 = 5 * 2 ^ (n + 1) - 6N + 2-12 + 12 = 5 * 2 ^ (n + 1) - 6N + 2
Right = - 2An + 10bn
=-2（3n-1）+10*2^n
=-2（3n-1）+5*2*2^n
=-6n+2+5*2^(n+1)
Immediate evidence

A = {x | y = x + 1} B = {x | y = root x + 2} C = {y | y = x ^ 2 + 1}, find anb; ANC; cub

(1) B = {x | y = root x + 2} that is x > = - 2
A = {x | y = x + 1} means that x belongs to R
AnB=X>=-2
(2) C = {y | y = x ^ 2 + 1} because x ^ 2 + 1 has a minimum value of 1 (at this time x = 0), so Y > = 1
AnC>=1
（3）CuB>=1

If the sequence {an} satisfies A1 > 0, A2 = 9 and has an + 1 = A1 * an for any positive integer n, find the general term formula of the sequence {an}

For any positive integer n, there is a (n + 1) = A1 * an
That is: a (n + 1) / an = A1
The common ratio q = A1
And A2 = A1 * q = q ^ 2 = 9
q＝3
an＝q^n＝3^n

For positive integers a and B, define that a △ B is equal to the sum of B consecutive positive integers starting from a, such as 2 △ 3 = 2 + 3 + 4, or 5 △ 4 = 5 + 6 + 7 + 8 = 26. If 1 △ x = 15, find X

From 2 △ 3 = 2 + 3 + 4, 5 △ 4 = 4 + 6 + 7 + 8 = 25, 1 △ x = 1 + 2 + 3 + There are x, ∵ 1 + 2 + 3 + 4 + 5 = 15 ∵ x = 5

All the multiples of positive integers of 9 add up to 9. Why? For example, 2 times 9 equals 18, 1 plus 8 equals 9

N * 9 = n * (10-1) = 10n-n = 10n-10 + 10-N = 10 (n-1) + 10-n. we should understand that the ten digit of 10A + B is a, and the single digit is B. the above problem: the number of ten digit is n-1, the number of single digit is 10-N, and the number of single digit plus ten digit = n-1 + 10-N = 9

Four mutually unequal integers ABCD, if ABCD = 9, find a + B + C + D

The four numbers are - 3, - 1,1,3
a+b+c+d=0

Factorization of mathematics a4-12a2b2 + B4 in Grade 8

Decomposition factor: A4 + 2a2b2 + B4

(a²+b²)²

Factorization of a2c2-b2c2-a4 + B4