The known set M = {x | x + 1 | + | x-3 | > 8}, P = {x | x ^ 2 + (A-8) x-8a

The known set M = {x | x + 1 | + | x-3 | > 8}, P = {x | x ^ 2 + (A-8) x-8a

In M = {{x {x {x 124124124124124124124\124\124124124124\124124\124\124\124\\124\124\124\\\\\124\124124124\124\\\\\124 {{5 {5 {5 {5 {5 {5 {5 {5 ≤ - a ≤ 5 (...)

If M = {1,3, X}, n = {x2,1}, and Mun = {1,3, X}, then the number of X satisfying the condition is ()
A.1 B.2 C.3 D.4

Three
0, positive and negative root sign three

Let m = {x | x ^ 2 + 2 (1-A) x + 3-a

Let y = x & # 178; + 2 (1-A) x + 3-A be regarded as a quadratic function. If its opening is upward, then y = x & # 178; + 2 (1-A) x + 3-A ≤ 0, the value of X is the value of X below the y-axis, that is, the part between the intersection of the quadratic function and the y-axis

How to solve this inequality when log2 (3x ^ 2-2x-5) is greater than or equal to log2 (4x ^ 2 + X-5)

∵ log [2] x increases monotonically in the domain
∴3x^2-2x-5≥4x^2+x-5,3x^2-2x-5>0,4x^2+x-5>0
0≤x≤3;x5/3;x1
∴5/3

Set a = {(x, y) | y = x2}. B = {(x, y) | y = x + 2}, where a intersects B equal to

It is equivalent to solving the equations
y=x^2
y=x+2
The solution is: x ^ 2 = x + 2,
(x-2)(x+1)=0
x=2,-1
y=4,1
So a intersection B = {(2,4), (- 1,1)}

Given a = {(x, y) | y = x2-1, X ∈ r}, B = {(x, y) | y = - x2 + 7, X ∈ r}, then a ∩ B=______ ．

From the equation system: y = x2 − 1y = − x2 + 7, the solution is x = 2Y = 3 or x = − 2Y = 3, then a ∩ B = {(2,3), (- 2,3)} so the answer is: {(2,3), (- 2,3)}

The known set a = {y | y = x2-1, x = R} a intersects B=
Given the set a = {y | y = x2-1, x = R}, B = {x | y = 3-x2}, then the intersection of a and B = {x | y = 3-x2}.

∵A={y|y=x^2-1,x=R}
∵y=x^2-1≥-1
∴A={y|y≥-1｝
∵B=｛x|y=√（3-x^2）｝
∴3-x^2≥0
-√3≤x≤√3
∴B=｛x|-√3≤x≤√3｝
∴A∩B=｛x│-1≤x≤√3｝

Let a = {x | x ^ 2 / 4, y ^ 2 / 3 = 1} and B {y | y ^ 2 / 3-x ^ 2 / 4 = 1}, then the intersection of a and B is equal
Let a = {x | x ^ 2 / 4 + y ^ 2 / 3 = 1} and B = {y | y ^ 2 / 3-x ^ 2 / 4 = 1}, then a and B are equal

A={x|-2≤x≤2}
B = {y | y ≤ - √ 3 or Y ≥ √ 3}
So a ∩ B = {x | - 2 ≤ x ≤ - √ 3 or √ 3 ≤ x ≤ 2}
Note: the representative elements in the collection

Given a = {(x, y) | X-Y = 3} and B = {(x, y) | 2x + y = 9}, then the intersection of a and B is equal to

4,1

Given that a = a ≤ x ≤ a + 3, B = x Ⅰ, X ＜ - 1 or X ＞ 5, the intersection of a and B is an empty set, the range of a is obtained. Second, a and B are equal to B, the range of a is obtained

a≥-1 a+3≤5
So - 1 ≤ a ≤ 2
A is not equal to B
Then a is a subset of B
A + 3 is less than - 1 or a > 5
A ＜ - 4 or a ＞ 5