For the nonempty subset a of the set M = {x | 1 ≤ x ≤ 10, X ∈ n}, let every element K in a be, 8. (2006 &; Shaoyang simulation, Hunan Province) the known set M = {x | 1 ≤ x ≤ 10, X ∈ n}, for its non empty subset a, multiply each element K in a by (- 1) k, and then sum (for example, a = {1,3,6}, the sum can be (- 1) &; 1 + (- 1) 3 &; 3 + (- 1) 6 &; 6 = 2), then for all non empty subsets of M, the sum of these sums is In all subsets of set M, 2 ^ 9 = 512 contain element 1. Similarly, it contains elements 2, 3 So for all the nonempty subsets of M, the sum of these sums is 2 ^ 9 (- 1 + 2-3 + 4 +...) -9+10)=5*2^9 (why does 2 ^ 9 = 512 contain element 1)

For the nonempty subset a of the set M = {x | 1 ≤ x ≤ 10, X ∈ n}, let every element K in a be, 8. (2006 &; Shaoyang simulation, Hunan Province) the known set M = {x | 1 ≤ x ≤ 10, X ∈ n}, for its non empty subset a, multiply each element K in a by (- 1) k, and then sum (for example, a = {1,3,6}, the sum can be (- 1) &; 1 + (- 1) 3 &; 3 + (- 1) 6 &; 6 = 2), then for all non empty subsets of M, the sum of these sums is In all subsets of set M, 2 ^ 9 = 512 contain element 1. Similarly, it contains elements 2, 3 So for all the nonempty subsets of M, the sum of these sums is 2 ^ 9 (- 1 + 2-3 + 4 +...) -9+10)=5*2^9 (why does 2 ^ 9 = 512 contain element 1)

Set M has 2 ^ 10-1 nonempty subsets. Let's first look at the nonempty subsets without 1. The nonempty subsets without 1 can be regarded as n = {x | 2 ≤ x ≤ 10, X ∈ n} nonempty subsets, because set n has 2 ^ 9-1 nonempty subsets, then all subsets of set m with 1 have (2 ^ 10-1) - (2 ^ 9-1) = 2 ^ 9

A={xIx=2k+1,k∈Z}B={xIx=2k-1,k∈Z}C={xIx=4k+1,k∈Z}
A. Is the relationship between B and C C C &; a = B?

Correct, a: 2K + 1 = 2 (K + 1) - 1
B：2k-1
K + 1 and K are equivalent under K ∈ Z, so a = B
C：4k+1=2（2k+1）-1
2K + 1 is less than K + 1 and K, so C belongs to a = B

Given the square of quadratic function y = x + (4K + 1) x + 2k-1
1. Prove that there must be two intersections between the parabola and the x-axis
2. If (x1,0) and (x2,0) are two intersections of parabola and X axis, and (x1-2) (x2-2) = 2k-3, find the value of K

1.
The square of X + (4K + 1) x + 2k-1 = 0
Discriminant = (4K + 1) ^ 2-4 (2k-1) = 16K ^ 2 + 5 > 0
So: the equation must have two real roots
So: there must be two intersections between the square + (4K + 1) x + 2k-1 of the parabola y = x and the X axis
two
x1+x2=-(4k+1)
x1x2=2k-1
And: (x1-2) (x2-2) = 2k-3
x1x1-2(x1+x2)+4=2k-3
(2k-1)+2(4k+1)+4=2k-3
k=-1