Let a = {y ｜ y = (x power of 3), X ∈ r}, B = {y ｜ y = 1 - (the square of x), X ∈ r}, then a ∩ B =?

Let a = {y ｜ y = (x power of 3), X ∈ r}, B = {y ｜ y = 1 - (the square of x), X ∈ r}, then a ∩ B =?

According to a set, Y > 0, and B set Y < = 1. Combined with two inequalities, we know that the intersection of a and B is
{0 if you don't know the solution of B set, the inequality 1-x ^ 2 = y can be transformed into x ^ 2 = 1-y, because x ^ 2 > = O, then 1-y > = 0, y < = 1

Given the set a = {x | X-1 > 0}, B = {y | y = 2 to the power of X}

Solve a = {x | x > 1} and solve the equation
B = {y | Y > 0} is equivalent to the value field
So a ∩ B is a real number greater than 1, {x | x > 1}

The set a = {y belongs to the power of X} of R | y = 2, and B = {- 1,0,1} needs to be resolved A. A intersection B = {0,1} B. A and B = (0, positive infinity) C. (CRA) intersection B = {- 1, 0}

Choose C. from the x power of y = 2, Y > 0

In mathematics, if set a is equal to y │ y = ln (1-x), and set B is equal to y │ y = x ^ 2, then a ∩ B =?

a. Is B a, B? If so, the solution of the equation ln (1-x) - XX = 0 is obtained

Let a = (x| x? - 4 < 0). B = (x | y = ln (x-1)), then a ∩ B =? Finding the intersection of a and B

Set a is the inequality x 2-4

Given the set a = Z, B = {X / y = ln (9-x ^ 2)}, find a ∩ B

B={x/y=ln(9-x^2)}=B={x/9-x^2>0}=B={x| -3

Given the real number a ≠ 0, the set a = {x | y = ln ax / x-a ^ 2-1} B = {x ｜ (X-2) (x-3a-1) 0）｝ (1) If a = 2, find the set a (2) Proposition p: X ∈ a, proposition q: X ∈ B, if P is a necessary condition of Q, find the value range of real number a

0

0

The graphs of the functions y = 3cos2 π X and y = 3x are shown in the figure,
So a = {x | 3cos2 π x = 3x, X ∈ r} = {x1, X2, 1}, B = {y | y2 = 1, y ∈ r} = {- 1, 1},
So a ∩ B = {x1, X2, 1} ∩ {- 1, 1} = {1}
So the answer is {1}

It is known that the set M = {x (x + 1) 2 ≤ 1}, P = {y y = 4 X-1 power + 1, X ∈ m, 3 / 4 ＜ a ≤ 1}, and the complete set u = R, Finding the complement set of M ∪ p

M = {x | 1} = {x | - 2}=

-2A (x + y) cubic × (- 3) a 2 (x + y) quartic

-2A (x + y) cubic × (- 3) a 2 (x + y) quartic
=【（-2）*（-3）】a^(1+2)*(x+y)^(3+4)
=6a^3(x+y)^7