1. Given the set a = {x | x2-3x + 2 = 0}, B = {1,2}, C = {x | x ＜ 9, X ∈ n}, fill in the blanks with appropriate symbols A__ B ,A C,{2} C ,2 C Fill in the blanks: 0 &;, {0} &;, &; {&;}, {x | x | 0} &; Fill in the blanks is the second question

1. Given the set a = {x | x2-3x + 2 = 0}, B = {1,2}, C = {x | x ＜ 9, X ∈ n}, fill in the blanks with appropriate symbols A__ B ,A C,{2} C ,2 C Fill in the blanks: 0 &;, {0} &;, &; {&;}, {x | x | 0} &; Fill in the blanks is the second question

1. X & # 178; - 3x + 2 = 0 (x-1) (X-2) = 0 x = 1 or x = 2x = 2 is the root of the equation x & # 178; + (a + 1) x + (A & # 178; - 14) = 0. X = 2 is substituted into the equation: 4 + 2 (a + 1) + A & # 178; - 14 = 0. When a & # 178; + 2a-8 = 0 (a + 4) (A-2) = 0 a = - 4 or a = 2A = - 4, the equation becomes X & # 178; - 3x + 2 = 0, which is the same equation as in a

Let a = {Y / y = x2 + 3x + 3}, B = {X / y = x2 + 3x + 3}, find (CUA) ∩ B

Function y = x & # 178; + 3x + 3 = (x + 3 / 2) & # 178; + 3 / 4
The set a represents the value range of the function, that is, the value range of Y [3 / 4, + ∞)
Set B represents the domain of function, that is, the value range of X is r
So (the complement of a) ∩ B = (- ∞, 3 / 4)

Given the set a = {x 3 ≤ x ＜ 7}, B = {x 2 ＜ x ＜ 10}, find Cr (a ∪ b), Cr (a ∩ b), (C
Given the set a = {x 3 ≤ x < 7}, B = {x 2 < x < 10}, find Cr (a ∪ b), Cr (a ∩ b), (CRA) ∩ B, a ∪ (CRB)
CR stands for complement. I don't know how to play

1. A ∪ B = {2 ＜ x ＜ 10}, R is a real number, that is Cr (a ∪ b) = {x ｜ x ≤ 2 or 10 ≤ x},
2. A ∩ B = {3 ≤ x ＜ 7}, R is a real number, that is Cr (a ∩ b) = {x ∩ x ＜ 3 or 7 ≤ x},
3. It can be seen from the above that a = a ∩ B, that is CRA = Cr (a ∩ b) = {x ∩ x ＜ 3 or 7 ≤ x},
So (CRA) ∩ B = {2 ＜ x ＜ 3 or 7 ≤ x ＜ 10}
4. Similarly, CRB = Cr (a ∪ b) = {x ∪ x ≤ 2 or 10 ≤ x}, a ∪ (CRB) = {x ∪ x ≤ 2 or 3 ≤ x < 7 or 10 ≤ x}

Let u = R, a = {x ︱ x is greater than or equal to 5 or X is less than - 7}, B = {x ︱ 8 is less than or equal to 12}
Finding CUA, cub, (CUA) Union (cub), (CUA) intersection (cub)

CuA={x｜-7≤x＜5}
Cub = = {x ︱ x greater than 12 or X less than or equal to - 8}
(CUA) Union (cub) = empty set
(CUA) intersection (cub) = {X - 7 ≤ x < 5x or x greater than 12 or X less than or equal to - 8}