If a belongs to a, then (1 / 1-A) belongs to a, and 1 does not belong to a.1). If 2 belongs to a, and a has three elements, find the set a 2) Verification: if a belongs to a, then (1-1 / a) belongs to a

If a belongs to a, then (1 / 1-A) belongs to a, and 1 does not belong to a.1). If 2 belongs to a, and a has three elements, find the set a 2) Verification: if a belongs to a, then (1-1 / a) belongs to a

Because 2 ∈ a, and 1 / (1-2) = - 1, so - 1 ∈ a
Taking - 1 into 1 / (1-A) yields: 1 / [1 - (- 1)] = 1 / 2, so 1 / 2 ∈ a
If 1 / 2 is brought into 1 / (1-A), 1 / (1-1 / 2) = 2
Therefore, the three elements of set a are: 2, - 1, 1 / 2
A={2、-1、1/2}

A set s whose elements are known to be real numbers satisfies the following conditions: (1) 1,0 does not belong to s; (2) if a ∈ s, then 1 △ (1-A) ∈ S. if {2, - 2} is contained in s, find the set s with the least number of elements

Because {2, - 2} is contained in s, so 2 ∈ s, and - 2 ∈ s
From 2 ∈ s, then 1 △ (1-2) = - 1 ∈ s, then 1 △ (1 - (- 1)) = 1 / 2 ∈ s, then 1 △ (1-1 / 2) = 2 ∈ s
From - 2 ∈ s, then 1 △ 1 - (- 2)) = 1 / 3 ∈ s, then 1 △ 1 / 3 = 3 / 2 ∈ s, then 1 △ 1-3 / 2 = - 2 ∈ s
So s = {2, - 2, - 1,1 / 2,1 / 3,3 / 2}

A set s whose elements are known to be real numbers satisfies the following conditions: (1) 1,0 does not belong to s; (2) if a ∈ s, then 1 △ (1-A) ∈ S. if {2, - 2} is contained in s, find the set s with the least number of elements

Because 2 ∈ s, so 1 △ 1-2 = - 1 ∈ s, so 1 △ 1 - (- 1) = 1 / 2 ∈ s, so 1 △ 1 - (1 / 2)] = 2 ∈ s, because - 2 ∈ s, so 1 △ 1 - (1 / 3) = 3 / 2 ∈ s, so 1 △ 1 - (3 / 2)] = - 2 ∈ s, so the set with the least number of elements s = {2, - 1,1 / 2, - 2,1 / 3

Small real numbers can form a set. Why not

How small is it? Is 1 small? 0.5 is smaller, 0.05 is smaller. The set needs to be determined. Can you determine a very small real number

The set of all real numbers greater than 3 can be expressed as 3,4,5, right

｛x│x＞3,x∈R}

If the proposition "x ∈ R, 2x2-3ax + 9 ＜ 0" is false, then the value range of real number a is ()
A. [-22，22]B. [-2，2]C. [-2，2]D. （-22，22）

If there is x ∈ R, such that 2x2-3ax + 9 ＜ 0 is a false proposition, then its negation is a true proposition, that is to say, ∀ x ∈ R, there are 2x2-3ax + 9 ≥ 0. According to the discussion of the solution of the quadratic inequality with one variable, we can see that the value range of △ = 9a2-72 ≤ 0, ∀ 22 ≤ a ≤ 22. A is [- 22, 22]

Let the set a = {x | - 4 ＜ x ＜ 2}, B = {x | - X & sup2; - 3ax + 2A & sup2; = 0, X ∈ r}, find the range of real number a of a that B contains

x²-3ax+2a²=0 (x-a)(x-2a)=0 x=a,x=2a
B is contained in a
(1)a>0,a

Set a = {x | - 2

x²-3ax+2a²=0
(x-a)(x-2a)=0
X = a or x = 2A
B is really contained in a
-2

The number of positive real roots of equation x2-5x + 2 + 2 / x = 0 is?

By drawing method:
F (x) = x ^ 2-5x + 2 = (X-5 / 2) ^ 2-17 / 4, the image is in quadrant 1,2,4
G (x) = - 2 / x, the image is in quadrant 2,4
There is an intersection in quadrant 2
There are two intersections in quadrant 4
Therefore, the original equation has three real roots

Proof: when m is a real number, at least one of the quadratic equation x2-5x + M = 0 and the equation 2x2 + x-6-m = 0 has a real root

Assuming that the above two equations have no real roots, then △ 1 = 25 − 4m ＜ 0, ① △ 2 = 1 + 4 × 2 × (6 + m) ＜ 0, ② ① get m ＞ 254, ② get m ＜ 498, such that m does not exist. At least one of the equations has a real root