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If u = R, a = {x | x2-x ＞ 0}, B = {x | LNX ≤ 0}, then (∁ UA) ∩ B = () A. （0，1]B. （-∞，0）∪（1，+∞）C. ∅D. （0，1）
From the inequality in a, we can get: X (x-1) > 0, we can get: x > 1 or x < 0, that is, a = (- ∞, 0) ∪ (1, + ∞), ∪ complete set u = R, ∁ UA = [0, 1]. From the inequality in B, we can get: LNX ≤ ln1, that is, 0 ∁ UA) ∩ B = (0, 1)
Given the set u = R, ∁ UA = {x | x2 + 6x ≠ 0}, B = {x | x2 + 3 (a + 1) x + A2-1 = 0} and a ∪ B = a, find the value range of real number a
CUA = CUA = {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 {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} x {x} x {x} x} X . To sum up, the value range of a is {a | a ＜ - 135, or a ≥ - 1, or a = 1 Or a = 17}
Let u = R, a = {x ∣ x is less than or equal to 1}, B = [x ∣ x is greater than a + 1}, and CUA is a subset of B
A = {x ∣ x is less than or equal to 1},
CuA={x|x>1}
CUA is a subset of B
a+1≤1
a≤0
The complete set u = R, {x ∣ - 1 is known
CUA = (- ∞, - 1] intersection [1, + ∞)
CUA intersection u = (- ∞, - 1] intersection [1, + ∞)
CUA and u = u
A intersection CUA = empty set
A and CUA = u
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