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Given the complete set u = {x | x ＜ = 4}, set a = {x | - 2 ＜ x ＜ 3}, B = {x | - 3 ＜ = x ＜ = 2}, find the intersection of (CUA} UB, a (cub) Wait online, hurry!
CuA={x|3
RELATED INFORMATIONS
- 1. A = {x | x ≤ - 3 or X ＞ 0} B = {x | - 4 ＜ x ≤ 1} u = R find anb (CUA) UB an (cub) （CuA)U(CuB) Cu(AUB)
- 2. Let u = R, a = {x | - 1
- 3. Set u = {x | x ≤ 10, and X ∈ positive integers}, A.B is the proper subset of u, and a ∩ B = {3,5}, (cub) ∩ a = {1,2,4}, (CUA) ∩ (cub) = {6,7} to find sets a and B
- 4. The complete set u = {x belongs to n * | X
- 5. Given the complete set u = {1,2,3,4,5,6,7,8,9,10}, the set a = {1,3,5,7}, B = {2,4,5,7}, then CUA ∩ cub=
- 6. Given the complete set u = R, set a = {x | x + 1 > = 0}, set B = {x | X's square-x-12 ∩ 0}, find (CUA) ∩ B
- 7. When u = {0,1,2,3,4,5,6,7,8}, a = {0,1,3,4,7}, B = {1,2}, verify that Cu (a ∪ b) = CUA ∩ cub and Cu (a ∩ b) = CUA ∪ cub There should be a detailed process, thank you
- 8. U = {1,2,3,4,5,6}, a = {2,3,5}, B = {1,4}, find Cu (AUB) and (CUA) ∩ (cub)
- 9. It is proved that Cu (a + b) = CUA + cub The answer should be detailed, the process should be standard, and it's better to understand. Few of us can understand what our math teacher said
- 10. For the dual rate proof of set, Cu (a ∩ b) = CUA ∪ cub, for the strict logic proof, no Venn diagram or enumeration
- 11. Given the complete set u = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10}, a ∩ (cub) = {1,3,5,7}, try to find the set B
- 12. Given the complete set u = R, set a = {x | 0 ＜ x ≤ 2}, B = {x | x ＜ - 3 or X ＞ 1}. (1) find a ∩ B. CUA (2) (CUA) ∪ (cub)
- 13. Let u = R, a = {x | x ＞ 3}, B = {x | 5 ＜ x ＜ 8}, find a ∩ B, a ∪ B, Cub ∪, a ∩ cub, CUA ∪ cub
- 14. Let u = R, a = {x | x > = 1}, B {x | 0
- 15. Let u = R, a = {x | x ≥ 1}, B = {x | 0 ＜ x ＜ 5}, find (∁ UA) ∪ B and a ∩ (∁ UB)
- 16. Three sets a = {X / x ^ 2-3x + 2 = 0}, B = {X / x ^ 2-ax + A-1 = 0}, C = {X / x ^ 2-bx + 2 = 0}, if a is the proper subset of B, AUC = a is the real number a, does B exist? If yes, find out a, B. if not, explain the reason B is the true subset of A, I don't quite understand
- 17. We know that the set a = {x belongs to R / ax ^ 2-3x + 2 = 0 The set a = {x belongs to R | ax ^ 2-3x + 2 = 0, a belongs to R}. 1. If a = empty set, find the value range of real number A. 2. If a is a single element set, find the value of real number a and a
- 18. Given that there is only one element in the set a = {x | (A-1) x ^ 2 + 3x-2 = 0}, we can find the value of real number a
- 19. Given that the set a = {x belongs to R | ax2-3x + 2 = 0} (1), if a = an empty set, find the value range of real number a (2) is a single element set, find the value of a and set a (3) Finding the set M = {a belongs to R | A is not equal to an empty set
- 20. Given the set a = {x ∈ R | X & sup2; + 3x + 3 = O}, B = {x ∈ R | X & sup2; - 5x + 6 = 0}, a is contained in P and true in B, find the set P satisfying such condition