The known set a = {x | X & sup2; - 2x-3 ≤ 0}, B = {x | X & sup2; - 2m + M & sup2; - 4 ≤ 0, X ∈ R, m ∈ r}} 1. If a ∩ B = {x | 0 ≤ x ≤ 3}, find the value of real number M 2. If a &; CRB, find the value range of real number M

The known set a = {x | X & sup2; - 2x-3 ≤ 0}, B = {x | X & sup2; - 2m + M & sup2; - 4 ≤ 0, X ∈ R, m ∈ r}} 1. If a ∩ B = {x | 0 ≤ x ≤ 3}, find the value of real number M 2. If a &; CRB, find the value range of real number M

Set a = {x | - 1 ≤ x ≤ 3}, set B = {x | - 2 ≤ x ≤ m + 2}
1. If a ∩ B = {x | 0 ≤ x ≤ 3}
Then m-2 = 0 and M = 2
2.CRB={x|xm+2}
If a &; CRB
Then 3M + 2
The solution is m > 5 or m

Given the complete set I = {2,3, a & # 178; - A + 2} and a = {a + 1,2}, CIA = {4}, then the real number a=

Complete set I = {2,3, a & # 178; - A + 2} and a = {a + 1,2}, CIA = {4}
Then I contains element 4, and a does not contain element 4, but contains elements 2 and 3
So a + 1 = 3 and a = 2
I hope I can help you. If you have any questions, please ask,

Given the complete set I = {- 1,2, a ^ 2-A}, a = {- 1, | a + 1}, CIA = {0}, find the value of A

Because CIA = {0},
So a ^ 2-A = 0, a = 0 or 1
|A + 1 | is not equal to 0, a is not equal to +, - 1
a=0

Given the complete set I = {- 1,2, a square - a}, a = {- 1, | a + 1}, CIA = {0}, find the value of A

According to the meaning of the topic, relative to the complete set I, the complement of a is 0, so a ^ 2-A = 0 = > A = 0 or 1
At the same time, a must have two elements that are the same as I, so the test can know that a = 1

Let the complete set I be equal to [2,3, the square of x plus 2x minus 3], a be equal to [5], and CIA be equal to [2, y]. Find the values of X and y
The inner CIA, the middle I, is a small one

I=A∪CIa
therefore
I={2,Y,5}
y=3
x^2+2x-3=5
X = - 4 or 2

Given the set a = {x | x | 1}, B = {x | x ＜ a}, if a ∪ B = a, then the value range of real number a is______ ．

∪ B = a, ∢ B ⊆ a ∪ a = {x | x | 1} = {x | x ＜ - 1 or X ＞ 1}, B = {x | x ＜ a}, ∢ a ≤ - 1, so the answer is: a ≤ - 1

Given the set a = {x | x ＜ a}, B = {x | 1 ＜ x ＜ 2}, and the complement of a ∪ B in R = R, then the value range of real number a is——
Where R is all real numbers. Note that it is the complement of a and B (that is, B takes the complement in R first, and then takes the union with a). Sorry, the complement symbol will not be typed. Don't misread it

Using the idea of combination of number and shape, draw the complement of B on the number axis, and then draw a, we can get the range of a is a ≥ 2

Given the set a = {x | x > 2}, B = {x | x > a}, if a is contained in B, then the value range of real number a
Is it ≥ 2 or ≤ 2?

a≤2
Yes, it's less than or equal to 2. You can see it by drawing. I changed the answer, so I'm on the last floor. Adopt me

If we know the set a = {x | x ≤ 1}, B = {x | x ＞ a}, and a ∪ B = R, then the value range of real number a is————
2 given the set a = {x | x ≤ 1}, B = {x | x ＞ a}, and a ∪ B = empty set, then the value range of real number a is————

1 is a ≤ 1, 2 is a ≥ 1

If the tangent of function f (x) = 1 / 3x ^ 3 + 1 / 2aX ^ 2 + (a ^ 2-8) x / 4 at x = 1 just passes through the function image, then a is equal to

Function f (x) = 1 / 3x ^ 3 + 1 / 2aX ^ 2 + 2bx extreme point,
So m, n are two of F '(x) = 0,
Since the image of y = f '(x) is a parabola with an opening upward,
And 0 & lt; M & lt; 1,1 & lt; n & lt; 2,
So f '(0) = 2B & gt; 0, f' (1) = a + 2B + 1 & gt; 0, f '(2) = 2A + 2B + 4 & gt; 0,
The above three formulas are the restrictions of a and B,
(b-2) / (A-1) is the slope of (a, b) and (1,2),
Draw constraints in rectangular coordinates
The range of (b-2) / (A-1) is (1 / 4,1)