Let f (x) = log2 (2x-3) be defined as set M, and G (x) = √ 3-x + √ x + 1 be defined as set n. let u = R Q: (1) . set M; n (2) . set (# 8705; UM) ∩ n, m ∪ n

Let f (x) = log2 (2x-3) be defined as set M, and G (x) = √ 3-x + √ x + 1 be defined as set n. let u = R Q: (1) . set M; n (2) . set (# 8705; UM) ∩ n, m ∪ n

（1）
The domain of function f (x) = log2 (2x-3) is set M,
The solution of 2x-3 > 0 is x > 3 / 2
∴M=(3/2,+∞)
The domain of the function g (x) = √ 3-x + √ x + 1 is set n
{3-x≥0
{x+1≥0
The solution is - 1 ≤ x ≤ 3
∴N=[-1,3]
(2)
CuM=(-∞,3/2]
(CuM)∩N=(-∞,3/2]∩[-1,3]=[-1,3/2]
MUN=(3/2,+∞)U[-1,3]=[-1,+∞)
Solution: (1) 2x-3 > 0, so m = {x | x > 3 / 2}, 3-x > = 0, x + 1 > = 0, so n = {x | - 1}=
Given the complete set u = R, the domain of function y = 1 x + 1 is set a, and the domain of function y = log2 (x + 2) is set B, then set (∁ UA) ∩ B = ()
A. （-2，-1）B. （-2，-1]C. （-∞，-2）D. （-1，+∞）
The domain of the definition of the function y = 1 is {{x \\\\\\124\124\124\124\\\124\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\8745; {x | x ＞ - 2} = {x | - 2 ＜ x ≤ - 1}
Given solution set a = (1,2), function y = log2 (x-a-1) (2a-x) (a ≠ 1), if B intersection a ≠ empty set, find the value range of A
Zero
Given the set a = {x | 3 ≤ x ≤ 7}, B = {x | 1 ≤ x ≤ 9}, find ① Cr (a ∩ b) ② a ∪ (CRB)
①A∩B={x|3≤x≤7}
So Cr (a ∩ b) = {x | x ＜ 3 or X ＞ 7}
② CRB = {x | x ＜ 1 or X ＞ 9}
A ∪ (CRB) = {x | x ＜ 1 or 3 ≤ x ≤ 7 or X ＞ 9}
{x | X & gt; 7 or X & lt; 3}, {x | X & gt; 9, X & lt; 1 or 3 ≤ x ≤ 7}
Given a = {x x + 3 + x-4 ≤ 9}, B {x x & sup2; + 2x-15 ≥ 0}, find a ∩ B, Cr (a ∪ b)
A = {x + 3 + x-4 ≤ 9} = {X - 4 ≤ x ≤ 5}, B = {x | x2-2x-15 ≥ 0} = {x | x ≥ 5 or X ≤ - 3}
A ∩ B = {x | - 4 ≤ x ≤ - 3 or x = 5}, and a ∪ B = R, Cr (a ∪ b) = Φ
Let u = R, a = {x-x-a ＜ 1}, B = {x + 1 / X-2 ≤ 2}, and a be included in Cub, then find the value range of real number a
A ={x|-1+a
Let u = R, a = {x | - 1 ≤ x ≤ 6}, B = {x | - A + 2 ≤ x ≤ 2A}. If B is included in the complement of a, then the value range of A
A={x|-1≤x≤6}=[1,6]
Then CUA = (- ∞, 1) ∪ (6, + ∞)
When B = &;, a + 2 > 2a, A4
So A4
A4
Let u = R, a = {x | - 1 ≤ x ≤ 6}, B = {x | - A + 2 ≤ x ≤ 2A}, if B is included in the complement of a, then the value range of a is questionable
A={x|-1≤x≤6}=[1,6]
Then CUA = (- ∞, 1) ∪ (6, + ∞)
When B = &;, a + 2 > 2a, A4
So A4
The question is marked with []. I really don't understand it. Please explain it
Because the complement of a is CUA = (- ∞, 1) ∪ (6, + ∞)
B is contained in CUA = (- ∞, 1) ∪ (6, + ∞)
So B must be all in (- ∞, 1) ∪ (6, + ∞)
So it's all in the first part, or it's all in the second part
If they are all in the previous part, the maximum value of B must be less than the maximum value of (- ∞, 1), that is, 2A6
When B is not equal to an empty set, there are two cases: if B is contained in (- ∞, 1), there will be 2A6.
If B belongs to the complement of a, it belongs to either the left segment or the right segment. If B belongs to the left segment, it is the rightmost segment whose maximum value is less than the left segment, that is, 2A
If a = 3, B = 5, C = 7, C = 120 ° in △ ABC, then Sina + Sina =?
cosA=(25+49-9)/70=13/14
A is an acute angle
sinA=3*3^0.5/14
Given that r = u, a = {- 1 ≤ x ≤ 4}, B = {a ≤ x ≤ 2a-2}, the complement of B is contained in a, the value range of a is obtained
Are you sure there is no problem in your problem narration?
(1) Is r = u or u = R? (2) "the complement of B contains a" or "the complement of B contains a"?
If u = R, no matter what the value of a is, the complement of B cannot be included in a
A ≤ 2a-2, because if B is an empty set, the complement of B will not be included in a, so a is greater than or equal to 2. Then, you draw a number axis, because a = {- 1 ≤ x ≤ 4}, so b set must be in a set, and ah, do you know the subset by your inclusion? If it's a subset, then only a is greater than or equal to 2, right? I'm just a senior one, please forgive me... ... unfold
A ≤ 2a-2, because if B is an empty set, the complement of B will not be included in a, so a is greater than or equal to 2. Then, you draw a number axis, because a = {- 1 ≤ x ≤ 4}, so b set must be in a set, and ah, do you know the subset by your inclusion? If it's a subset, then only a is greater than or equal to 2, right? I'm just a senior one, please forgive me... Put it away