Given that a and B are two unequal real numbers, set M = {a2-4a, - 1}, n = {b2-4b + 1, - 2}, F: X → x means that if the element X in M is mapped to the set N and is still x, then a + B is equal to______ ．

Given that a and B are two unequal real numbers, set M = {a2-4a, - 1}, n = {b2-4b + 1, - 2}, F: X → x means that if the element X in M is mapped to the set N and is still x, then a + B is equal to______ ．

∵ F: X → x means that the element X in M is mapped to the set n, which is still x, ∵ M = n, and ∵ set M = {a2-4a, - 1}, n = {b2-4b + 1, - 2}, ∵ a2-4a = - 2, and b2-4b + 1 = - 1, that is, a and B are the two roots of equation x2-4x + 2 = 0, so a + B = 4, so the answer is: 4

Given the set P = {x | x = A2 + 4A + 1, a belongs to R}, q = {y | y = - B2 + 2B + 3, B belongs to R}, find the intersection Q of P=_ The complement of P and q =
A2 means the square of a, B2 means the square of B, thank you!

x=a^2+4a+1=(a+2)^2-3
So vertices (- 2, - 3)
Because the opening of the image is upward, x > = - 3
Because y = - B2 + 2B + 3 = - (B-1) ^ 2 + 4
So the vertex is (1,4)
Because the opening of the image is downward, so y

Given that the complete set is u, P = {x x = A & sup2; + 4A + 1, a ∈ r}, q = {y y = - B & sup2; + 2B + 3, B ∈ r}, find P ∩ Q

P={x｜x=(a+2)^2-3},P={x｜x>=-3};
Q={y｜y=-(b-1)^2+4},Q={y｜y

Given the set P = {x x = A & sup2; + 4A + 1, a ∈ r}, q = {X - B & sup2; + 2B + 3, B ∈ r}, find P ∩ Q

x=a²+4a+1=（a+2）²-3 ∴x≥-3
x=-b²+2b+3=-（b-1）²+4 ∴x≤4
∴P∩Q=={x｜-3≤x≤4}
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Given the set P = {x x = a ^ 2 + 4A + 1, a ∈ r}, q = {y = - B ^ 2 + 2B + 3, B ∈ r}, find P ∩ Q, P ∪ (the complement of Q in real number)

The range of the function represented by P and Q, then: P = {x | x ≥ - 3}, q = {y | y ≤ 4}
1、P∩Q={x|－3≤x≤4}；
2. If the complement of Q in real number set is {y | Y > 4}, the answer is: {x | x ≥ - 3}
Note: it is better to use the number axis to solve the problem

If the set a = {y | y = a ^ 2-4a + 3, a ∈ r}, and the set P = {x | x = 2 + 2b-b ^ 2, B ∈ r}, then m ∩ p=

M∩P={x|-1≤x≤3},
[^ 2 means square]
Vertex coordinates of a ^ 2-4a + 3:
(4ac-b^2)/4a=(4×1×3-16)/4=-1
-Vertex coordinates of B ^ 2 + 2B + 2:
(4ac-b^2)/4a=[4×(-1)×2-4]/(-4)=(-12)/(-4)=3
M∩P={x|-1≤x≤3}

Let a = {2, 8, a}, B = {2, a2-3a + 4}, and a ⊋ B, find the value of A

Because AB, so a2-3a + 4 = 8 or a2-3a + 4 = A. from a2-3a + 4 = 8, get a = 4 or a = - 1; from a2-3a + 4 = a, get a = 2. The test: when a = 2, the elements in the set a and B have repetition, which is contradictory with the set elements, so the value of a in line with the meaning of the question is - 1, 4

Set a = {the square of x-6x + 8 ＜ 0} B = {x (x-a) (x-3a) ＜ 0. If a is a proper subset of B, find a
If a intersects B = empty set, find a
If a intersects B = {x 3}

A = {the square of X - 6x + 8 ＜ 0} = {x 2 ＜ x 4};
B = {x (x-a) (x-3a) < 0} = {3A < x < A, a < 0} or empty set, a = 0 or {x a < x < 3a, a > 0}
(1) Because a is the proper subset of B, B = {x a < x 3a, a > 0} and a < 2,4 < 3A get 4 / 3 < a < 2
(2) A = 0, B = empty set, a intersection, B = empty set
A ＜ 0, B = {3a ＜ x ＜ a, a ＜ 0} so the element in B is negative, so B intersection a = empty set
A > 0, B = {x a < x 3a, a > 0} if a intersects B = empty set
If 3a ＜ = 2 or a ＞ = 4, then 0 ＜ a ＜ 2 / 3 or a ＞ = 4
So a ∈ (- ∞, 2 / 3) ∪ [4, + ∞)
(3) A intersection B = {x 3}

Let a = {2,8, a} B = {2, a ^ 2-3a + 4}, a contain B, and find the value of A
Is this a discussion of whether B is an empty set

A ^ 2-3a + 4 can only be 8 or a (2 already exists, should not be repeated) when a ^ 2-3a + 4 = 8, a = - 1 or 4A ^ 2-3a + 4 = a, a = 2, then the element has been repeated. I think the answer is - 1 or 4

If a belongs to R, then the set of all the values of a in a = | a, A2 + 3a | is____

a²+3a≠a
a²+2a≠0
a(a+2)≠0
A ≠ 0 or a ≠ - 2
So a = {AI a ∈ (- ∞, - 2) ∪ (- 2, 0) ∪ (0, + ∞)}