The complete set s = {1,2, x-2x}, a = {1, | 2x-1}, if CSA = {0}, then such a real number Given the complete set s = {1,2, x-2x}, a = {1, | 2x-1}, if CSA = {0}, does such real number x exist? If so, please explain the reason

The complete set s = {1,2, x-2x}, a = {1, | 2x-1}, if CSA = {0}, then such a real number Given the complete set s = {1,2, x-2x}, a = {1, | 2x-1}, if CSA = {0}, does such real number x exist? If so, please explain the reason

CsA=｛0
There are 0 elements in S
X quadratic - 2x solvable to
X=2 X=0
Substitute the value of X into B to test
|2x-1 | not equal to 2
non-existent
Let x ^ 2 be the power of X
We get x ^ 2-2x = 0, | 2x-1 | = 2 from the meaning of the title, and we get that X has no solution, so it doesn't exist
3-2x power of 2
The right side of the original formula is equal to the 4-3x power of 2, so 3-2x is less than 4-3x, that is, X is less than 1, so the value range of X is less than 1
2x-3>3x-4
X
We know three sets {x 2 = - x 2} a = {x 2 + x 2}
If B is a proper subset of a, a ∪ C = a, and the empty set is a proper subset of C, do the real numbers a and B exist? If so, find out the values of a and B; if not, explain the reason
Pro talk about the popular point of Europe, if you can tell me the solution of similar problems, plus wealth yo!
A: (x-1) (X-2) = 0A = {1,2} B is a proper subset of a, which indicates that B = {1} or {2} a ∪ C = a, and an empty set is a proper subset of C, which means that C is nonempty and a subset of a, that is, C = {1} or {2} or {1,2}
Great Satan: what are you talking about?
Given the set a = {x | x2-3x + 2 = 0}, C = {x | x2-x + 2m = 0}, if a ∩ C = C, find the value range of M
∵ a = {x | x2-3x + 2 = 0}, ∵ a = {1, 2}; ∵ C = {x | x2-x + 2m = 0}, and a ∩ C = C, so C ⊆ a; ① when C = Φ, △ = 1-8m ＜ 0, i.e. m ＞ 18; ② when C ≠ Φ, if C ⊊ a, obviously not true; if C = a, obviously not true; in conclusion, m ＞ 18
Given that the definition field of function f (x) is r, for any real number x, y, f (x + y) = f (x) + F (y) + 1 / 2 and f (1 / 2) = 0, if x > 1 / 2, f (x) > 0, monotonicity is obtained
Let X1 > x2 = > X1 - x2 > 0f (x1) - f (x2) = f (x1 - x2 + x2) - f (x2) = f (x1 - x2) + F (x2) + 1 / 2 - f (x2) = f (x1 - x2) + 1 / 2 when X1 - x2 > 1 / 2, f (x1 - x2) > 0, so f (x1) - f (x2) > 1 / 2 > 0 when X1 - x2 > 1 / 2, f (x1 - x2) > 1 / 2 > 0 when X1 - x2 < 1 / 2
Given that the image of function y = f (x) passes through point (1,0), and f (x) = (x ^ 2) - x + B, the first n terms of sequence {an} and Sn = f (n), n ∈ n +, find the general term formula of {an}
I just can't figure out what the 1 in (1,0) is going to be, X-1 = 1 or x = 1
This is where we want to enter the f (x) x (x) (x = 0) for the purpose of (x) x (x) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\theresults show that A1 = 2-2 = 0, also satisfy
Let f (x) be defined as r if x 1 and for any real number x, y ∈ R, f (x + y) = f (x) f (y)
(3) The sequence {an} satisfies A1 = f (0), and f (an + 1) = 1 / F
(-2-an)(n∈n*)
1: Find the general formula of [an}
2 when a > 1, the inequality (1 / an + 1) + (1 / an + 2) +. + 1 / a 2n > 12 / 35 (log (a + 1) x-log ax + 1) holds for all positive integers not less than 2, and the value range of X is obtained
The correct solution can increase the reward
Easyf (x + y) = f (x) f (y) f (0) = f (0) * f (0) f (0) = 0 or 1, if f (0) = 0f (x + y) = f (x) f (y) let x = 0f (y) = 0, f (x) is a constant function, which is not consistent with the case of x1. F (0) can only be 1A1 = 1 f (a (n + 1)) = 1 / F (- 2-An) f (a (n + 1)) f (- 2-An) = f (0) set f (x + y) = f (x) f (y
(1) Let x = 0, then f (0 + y) = f (0) * f (y), that is, f (y) = f (0) * f (y), f (0) = 1
(2) The contradiction between X1 = f (0) and proof.
It is known that the sum of the first n terms of the sequence {an} is SN. For all positive integers, the point (n, Sn) is on the image of the function f (x) = 2 ^ (x + 2) - 4. 1 find the general term formula. 2 let BN = an × log2an find the sum of the first n terms of BN
1、
According to the meaning of the title, we can draw a conclusion
Sn=2^(n+2)-4=4(2^n-1)
a1=S1=4(2^1-1)=4
an=Sn-Sn-1=4(2^n-1)-4[2^(n-1)]=4[2^n-2^(n-1)]=2(2*2^n-2^n)=2^(n+1)
The same holds for n = 1
The general formula of {an} is an = 2 ^ (n + 1)
Two
bn=anlog2(an)
=2^(n+1)log2[2^(n+1)]
=(n+1)2^(n+1)
Tn=b1+b2+...+bn=2*2^2+3*2^3+...+n2^n+(n+1)2^(n+1)
Tn/2=2*2+3*2^2+4*2^3+...+(n+1)2^n
Tn/2-Tn=2*2+2^2+2^3+...+2^n-(n+1)2^(n+1)
=2+2+2^2+2^3+...+2^n-(n+1)2^(n+1)
=2+2(2^n-1)/(2-1)-(n+1)2^(n+1)
=2+2^(n+1)-2-n2^(n+1)-2^(n+1)
=-n2^(n+1)
Tn/2=n2^(n+1)
Tn=n*2^(n+2)
(n, Sn) on the image of function f (x) = 2 ^ (x + 2) - 4
sn=2^(n+2)-4
s(n-1)=2^(n-1+2)-4=2^(n+1)-4
So A1 = S1 = 2 ^ (1 + 2) - 4 = 4
an=sn-s（n-1）【a≥2】
=2^(n+2)-4-[2^(n+1)-4]
=2^(n+2)-2^(n+1)
=2^(n+1)
BN = an × log2a... Expansion
(n, Sn) on the image of function f (x) = 2 ^ (x + 2) - 4
sn=2^(n+2)-4
s(n-1)=2^(n-1+2)-4=2^(n+1)-4
So A1 = S1 = 2 ^ (1 + 2) - 4 = 4
an=sn-s（n-1）【a≥2】
=2^(n+2)-4-[2^(n+1)-4]
=2^(n+2)-2^(n+1)
=2^(n+1)
bn=an×log2an
=2^(n+1)×log2 2^(n+1)
=2^(2n+2)
=(2^2)^(n+1)
=4^(n+1)
b1=4^(1+1)=16
Tn=b1[1-q^(n+1)]/(1-q)
=16*[1-4^(n+1)]/(1-4)
=[16-4^(n+3)]/(-3)
=4 ^ (n + 3) / 3-16 / 3
One
Sn=2^(n+2)-4
S(n-1)=2^(n+1)-4
an=Sn-S(n-1)=2^(n+1)
Two
BN = 2 ^ (n + 1) * (n + 1)
=（1-n）*2^(n+2)-4+2^(n+2)-2
=(2-n)^2^(n+2)-6
If the definition field of function f (x) is r, for any real number a, B, f (a + b) = f (a) f (b)
(1) Let f (1) = K (k is not equal to 0) find f (10)
(let x 1 solve the inequality f (x + 5) > 1 / F (x)
Because f (a + b) = f (a) f (b), and f (1) = K
So:
f(10)=f(1+9)=f(1)f(9)=kf(9)
=kf(1+8)=kf(1)f(8)=(k^2)f(8)
=……
=k^10
One. f(1+9)=f(1)f(9)
f(1+8)=f(1)f(8)
......
f(1+1)=f(1)f(1)
f(10)=k^10
Two.
From F (a + b) = f (a) f (b) and the conclusion in 1, f (5) = k ^ 5
f(x+5)=f(5)*f(x)=k^5*f(x)
The inequality f (x + 5) > 1 / F (x) can be reduced to
k^5*f(x)>1/f(x)
f(x)*f(x)>k^5
The rest of the meeting.
X1, f (x) > 0
1 f (1 + 1) = f (1) f (1) = the square of K, and so on, f (10) = the tenth power of K
And so on, f (1) + F (2) + F (1) = f (1) + F (2) = 10
Let the sum of the first n terms of the sequence {an} be SN. For all n ∈ n *, the point (n, Sn / N) is on the image of the function f (x) = x + an / 2x
Let the sum of the first n terms of the sequence {an} be SN. For all n ∈ n *, the point (n, Sn / N) is on the image of the function f (x) = x + (an) / 2x
Substituting the function (2x, Sn) = x / n
Sn=n^2+an/2
S(n+1)=(n+1)^2+a(n+1)/2
Subtraction of two formulas
2n+1=(an+a(n+1))/2
Then 2n + 3 = (a (n + 1) + a (n + 2)) / 2
Subtraction of two formulas
a(n+2)-an=4
And - A1 = 2, - A1 / 2
And S2 = a1 + A2 = 2 + A2 = 2 ^ 2 + A2 / 2, -- > A2 = 4
So when n is odd, an = a1 + (n-1) * 4 / 2 = 2n
When n is even, an = A2 + (n-2) / 2 * 4 = 2n
In conclusion, an = 2n