![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
What is Cr (a ∩ b) equal to if u = R, a {x ∩ x ＜ 0 or X ＞ 2 =, B = {X - 1 ＜ x ＜ 3 =? Please tell me the process, thank you
Complete set u = R, set a {x ｜ x ＜ 0 or X ＞ 2}, B = {x ｜ - 1 ＜ x ＜ 3},
∴(A∩B)={x｜-1
The complete set u = R, set a = {X / X less than or equal to A-1}, set B = {X / x greater than A-2}, set C = {X / X less than 0 or x greater than or equal to 4}
If the complement of a and B belongs to C, find the value range of real number a
∵ a = {x | x ≤ A-1}, B = {x | x ＞ A-2}, C = {x | x ＜ 0, or X ≥ 4}
A ∪ B = {x | A-2 ＜ x, or X ≤ A-1},
And ∵ for any a belonging to the complete set R, A-2 is always less than A-1
A ∪ B = {x | A-2 ＜ x, or X ≤ A-1} = R
The complements of (a ∪ b) = &;
The complement of ∪ B belongs to C, and an empty set is a subset of any set
The value range of a is the complete set R
RELATED INFORMATIONS
- 1. Given the complete set u = R, set a = {a | a ≥ 2, or a ≤ - 2}, B = {a | the equation AX ^ 2-x + 1 = 0 of X has real roots}, find a ∪ B, a ∪ (cub)
- 2. Given that the set u = x is greater than or equal to 2, the set a = y 3 is less than or equal to y less than 4, the set B = Z 2 is less than or equal to Z less than 5, find the intersection B of a's complement sets and the union a of B's complement sets
- 3. For real numbers x and y, a new operation *, X * y = ax + bx-3, where a and B are constants, if 1 * 2 = 9, (- 3) * 3 = 6, find the value of 2 * (- 7)
- 4. Given * * a = {(x, y) | ax + y = 1}, B = {(x, y) | x + ay = 1}, C = {(x, y) | X & # 178; + Y & # 178; = 1} then (1) What is the value of a when (a ∪ b) ∩ C is a 2-ary set (2) What is the value of a when (a ∪ b) ∩ C is a 3-ary set The answer is (1) a = 0 or 1 (2) a = - 1 ± √ 2, And I'm just a freshman,
- 5. Given ax + ay = 3, BX + by = 5, find [(a) 2 + (b) 2] [(x) 2 + (y) 2] 2 is the square
- 6. (ax+by)^2+(ay-bx)^2+2(ax+by)(ay-by)
- 7. Let a, B, x, y belong to R, and a ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 1, prove that the absolute value of AX + by is less than or equal to 1 In addition to using the mean inequality, there are other solutions
- 8. F (x) = ax + 1 / a (1-x), where a is greater than 0, let g (a) be the minimum value of F (x) when 0 is less than or equal to X and less than or equal to 1 (1) Find the analytic expression of G (a); (2) find the maximum value of G (a) f(x)=ax+(1-x)/a
- 9. The square plus ax minus B of the quadratic equation x with real coefficients is equal to zero, one root is between (0,1), one root is between (1,2), and a (0,2) 2) P (a, b), find the value range of the module of 0A vector multiplied by 0P vector △ 0P vector
- 10. Decomposition factor: (AX by) ^ 3 + (by CZ) ^ 3 + (CZ ax) ^ 3
- 11. 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
- 12. Let a, B, C be real numbers, f (x) = (x + a) (x2 + BX + C), G (x) = (AX + 1) (CX2 + BX + 1). Let s = {x | f (x) = 0, X ∈ r}, t = {x | g (x) = 0, X ∈ r}. If {s}, {t} are the number of elements of S, T & nbsp; respectively, then the following conclusion is impossible () A. {s} = 1 and {t} = 0b. {s} = 1 and {t} = 1C. {s} = 2 and {t} = 2D. {s} = 2 and {t} = 3
- 13. If the complete set u is r, a = {x | 1 ≤ x ≤ 2}, if the complement of B and a in R = R, the complement of B intersection a in R = {x | 0}
- 14. The known set a = {x | ax + BX + C > 0, a is not equal to 0} = {x | M
- 15. If inequality LG (2aX) / LG (a + x)
- 16. On the inequality 2x-m of X less than or equal to 0, the positive solution of X is 1.2.3, and the value range of M is obtained
- 17. If the solution set of inequality ax & # 178; + BX + C ＞ 0 (a ≠ 0) is an empty set, then the condition that coefficients a, B, C should satisfy is
- 18. It is known that the solution set of quadratic inequality ax ^ 2 + BX + C > 0 is {x | - 1
- 19. We know the proposition p: "for any x ∈ R, there exists m ∈ R, 4 ^ x + 2 ^ XM + 1 = 0". If the proposition is not p and is a false proposition, then the value of M of real number Δ > = 0, the solution is m = 2, why m > = 2 is not good
- 20. Given that x is a real number, find the value range of (x2 + 2x + 3) / (x2 + 2x + 1)