Let P.Q be two nonempty sets of real numbers. Define P + q = {x = │ x = - A + B, a ∈ P, B ∈ Q} if P = {0,1,2} Let P.Q be two nonempty sets of real numbers, and define P + q = {x = │ x = - A + B, a ∈ P, B ∈ Q}. If P = {0,1,2}, q = {- 1,1,6}, the sum of all elements of P + Q is -- Is it to calculate the elements one by one and then sum them up? Is there a simpler way?

Let P.Q be two nonempty sets of real numbers. Define P + q = {x = │ x = - A + B, a ∈ P, B ∈ Q} if P = {0,1,2} Let P.Q be two nonempty sets of real numbers, and define P + q = {x = │ x = - A + B, a ∈ P, B ∈ Q}. If P = {0,1,2}, q = {- 1,1,6}, the sum of all elements of P + Q is -- Is it to calculate the elements one by one and then sum them up? Is there a simpler way?

-a=0,-1,-2 b=-1,1,6
When - a = 0, x = - A + B = 0-1 / 0 + 1 / 0 + 6 = - 1 / 1 / 6
When - a = - 1, x = - A + B = - 1-1 / - 1 + 1 / - 1 + 6 = - 2 / 0 / 5
When - a = - 2, x = - A + B = - 2-1 / - 2 + 1 / - 2 + 6 = - 3 / - 1 / 4
So p + q = - 1 / 1 / 6 / - 2 / 0 / 5 / - 3 / - 1 / 4
So the sum of all elements of P + Q is 9

Given the set a = {x | x ≥ 4}, the domain of G (x) = 1 / √ 1-x + A is B. If a ∩ B = empty set;, what is the range of real number a?
Don't copy the answers, try to come up with a detailed process. Finally, don't make mistakes. This is to be written in the exercise book

Set B = {x | x

If the definition field of function f (x) = root x ^ 2 + ax + 1 is real number set R, then the value range of real number a is

It can be seen from the meaning that for any real number x, x ^ 2 + ax + 1 is always greater than or equal to 0, so the discriminant of root is less than or equal to 0
That is, a ^ 2 - 4

Given that the set P = {1,3}, the set Q = {x | MX-1 = 0}, if the set Q is a subset of the set P, then the value range of the real number m is?
The answer is {0,1,1 / 3}, but Q where MX-1 = 0, MX = 1, x = 1 / m, the molecule can't be 0, why the value range is 0

1) If x = 1, then M = 1
2) If x = 3, then M = 1 / 3
3) If q is an empty set, then M = 0, then q is also a subset of P