A = {a + 2, (a + 1) 2, A2 + 3A + 3}, B = {a + B, 1, A-B + 5}. If a = B, find the value of real numbers a and B

A = {a + 2, (a + 1) 2, A2 + 3A + 3}, B = {a + B, 1, A-B + 5}. If a = B, find the value of real numbers a and B

If a + 2 = 1, then a = - 1, then a ^ 2 + 3A + 3 = 1, which is contradictory to the anisotropy of elements
If (a + 1) ^ 2 = 1, then a = 0 or - 2:
1) When a = 0, a = {1,2,3}, B = 2, - B + 5 = 3; or B = 3, - B + 5 = 2
The solution is b = 2 or 3
2) When a = - 2, a ^ 2 + 3A + 3 = 1
If a ^ 2 + 3A + 3 = 1, then a ^ 2 + 3A + 2 = 0, a ≠ - 1, - 2, no solution
In conclusion, a = 0, B = 2 or 3

A = {x | x2 + 4x = 0}, B = {x | x2 + 2 (a + 1) x + A2-1 = 0} where a belongs to R, if B is a subset of a, find the value range of real number a+
A = {x | x2 + 4x = 0}, B = {x | x2 + 2 (a + 1) x + A2-1 = 0}, where a belongs to R, if B is a subset of a, the value range of real number a is obtained
What does - 2 (a + 1) = - 4 mean

The solution of the equation x ^ 2 + 4x = 0 is x = 0 or x = - 4
A = {x | x ^ 2 + 4x = 0} = {x | x = 0 or x = - 4} = {0, - 4}
B={x|x^2+2(a+1)x+a^2-1=0}
B is required to be included in a, because the element of B is also the solution of quadratic equation with one variable, so there are at most two elements in B
Obviously, when the equation in B has two different solutions, the two solutions must be 0 and - 4, otherwise B cannot be included in A. at this time, the equations in a and B are equivalent equations
-2 (a + 1) = - 4 + 0 = - 4 and a ^ 2-1 = 0, we can get a = 1 (that is, Veda's theorem)
When the two equations in B are equal, that is, there is only one solution, the discriminant = 0, that is, a + 1 = 0, a = - 1. At this time, the root of the equation is x = 0, that is, B = {0}. Obviously, B is also included in a, that is, a = - 1 also satisfies the condition
When the equation in B has no real solution, i.e. B is an empty set, B is also included in A. in this case, the discriminant ＜ 0 is required, i.e. a + 1 ＜ 0, a ＜ - 1
To sum up, the value range of the real number a satisfying the condition is: a ≤ - 1, or a = 1

Let a = {x | x2 + 4x = 0}, B = {x | x2 + 2 (a + 1) x + A2-1 = 0}, where x ∈ R, if B ⊆ a, find the value range of real number a

(a) if B = {{x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {x {0, - 4}}, {{B} {B} {B} {}} {{4 {4} 4) 2 (4) 2-8 (a + 1) 8 (a + 1) + 8 (a + 1) 8 (a + 1) 8 (a + 1) + this is a = 1, when the equation set has no solution. There is no solution. This is no solution. This is a = 1. This is no solution. This is a = 1. To sum up, the real number a = 1 & nbsp; or a ≤ - 1

Given the set M = {the square of XLX + 2x-a = 0}, if the empty set is the proper subset of M, the value range of real number a is obtained
I want to ask him to solve the problem
Solution: because m is the solution set of equation... (the above equation), the empty set is the proper subset of M,
So the equation has a real solution,
So △ is greater than or equal to 0, 4 + 4a is greater than or equal to 0,
So a is greater than or equal to - 1
Why?
M is the solution set of equation... (the above equation), and the empty set is the proper subset of M,
So the equation has a real solution?

Proper subsets of an empty set M
So m is not an empty set
Here the element M is x, which is the solution of the equation
If the equation has no solution, then there is no element in M, which is an empty set
So the equation has a solution

A set is x greater than 2A but less than a + 3, b set is x less than - 1 and more than 5. If a intersects B and is equal to an empty set, the value range of a is obtained

In order to make the two empty
It's only possible
The range of a is (2a, a + 3)
B is X5
And
-1

Given that set a = {x | x | x is less than or equal to - 1 or X is greater than or equal to 2}, set B = {x | (x-a) (x-a-5) is less than 0} if B is really contained in a, find the real number
It is known that set a = {x | x | x is less than or equal to - 1 or X is greater than or equal to 2}, and set B = {x | (x-a) (x-a-5) is less than 0}
If B is really included in a, find the value range of real number a
If the intersection of a and B is equal to R, find the value range of real number a

1. A = {x | x ≤ - 1 or X ≥ 2} (x-a) (x-a-5) ＜ 0, the solution is a ＜ x ＜ a + 5, that is, B = {x | a ＜ x ＜ a + 5} because B is really contained in a, so a ≥ 2 or a + 5 ≤ - 1, the solution is a ≥ 2 or a ≤ - 6, that is, the value range of a is a ≥ 2 or a ≤ - 62, the intersection of a is b = R, so a ≤ - 1 and a + 5 ≥ 2, that is - 3 ≤ a ≤ - 1, so the value range of a is - 3 ≤ a

Given that the set a = {x | ax's square + X + 1 = 0, X belongs to R}, and the intersection of a {x | x is greater than or equal to 0} = empty set, find the value range of real number a

That is to say, X in a has no solution greater than or equal to zero. (if there is a solution greater than or equal to 0, the intersection of {X / X ≥ 0} is not empty) ax & # 178; + X + 1 = 0, Δ = 1-4ax = (- 1 ± √ Δ) / 2A. So there are three conditions about a: (1) ax & # 178; + X + 1 = 0 = > a ≠ 0 (2) Δ = 1-4a = > 1-4a > 0 = > A (- 1

Given that the set a = {x a × + B = 1} B = {x ax-b > 4} A is not equal to 0, if every element in a must be the element in B, the value range of real number B is obtained
Speed, high school mathematics

A = {(1-B) / a} B = {x | x > (4 + b) / a} so if an element in a is an element in B, then (1-B) / a > (4 + b) / a (1) if a > 0, then the range of B is B

Set a = {x | 1 ＜ ax ＜ 2}, set B = {x | - 1 ＜ x ＜ 1}. If a & # 8838; B, find the planting range of real number a

When a > 0, a = {x | 1 / A-1, a

M = {x | x-a = 0}, n = {x | AX-1 = 0}, if n &; m, find the value set of real number a

M = {x | x-a = 0}, then M = {a}. From m ∩ n = n, m contains n
1. N = &;, that is, a = 0
2. N = m, there is a = 1 / A, that is a = 1 or - 1