As shown in the figure, it is known that the image of the first-order function y = KX + B passes through the point P (3,2) and intersects the image of the inverse scale function y = 2x (x > 0) at the point Q (m, n). When the value of the first-order function y increases with the increase of the value of X, the value range of M is______ ．

As shown in the figure, it is known that the image of the first-order function y = KX + B passes through the point P (3,2) and intersects the image of the inverse scale function y = 2x (x > 0) at the point Q (m, n). When the value of the first-order function y increases with the increase of the value of X, the value range of M is______ ．

As shown in the figure, substituting y = 2 into y = 2x to get x = 1; substituting x = 3 into y = 2x to get y = 23, so the coordinates of point a and point B are (1,2) and (3,23). Because the value of primary function y increases with the increase of x value, point Q can only be between point a and point B, so the value range of M is 1 < m < 3 ＜m＜3．

Let m = {x | - 1

Draw the value range of the number axis X, and then simplify n, X is less than or equal to K, so K is greater than - 1
Or do you think so or draw the number axis? Because m intersection n is not an empty set, so x is less than or equal to K. on the number axis, there must be an inclusion relationship with - 1,2, so K must be greater than the minimum value of M. if k = - 1, then M intersection n = empty, so K is greater than - 1

Given the set a = {x | - 3 ≤ x ≤ 4}, B = {x | - 2m-1 ≤ x ≤ m + 1}, and a ∩ B = B, find the value range of real number M

∵A∩B=B⇔B⊆A… (2 points) (1) B = ∞, that is, 2m-1 ＞ m + 1 ⇔ m ＞ 2 (4 points) (2) when B ≠ 0, B ⊆ a ⇔ 2m − 1 ≤ m + 12m − 1 ≥ − 3M + 1 ≤ 4 ⇔ m ≤ 2m ≥ − 1m ≤ 3, ⇔ - 1 ≤ m ≤ 2 In conclusion, the range of M is [- 1, + ∞) (8 points)

Given the set M = {0,1,2}, n = {x | x = 2A, a ∈ m}, then the set M ∩ n = () how can the pile after n be calculated to be 0.2.4?
How to calculate the result for a master? I know the process

X = 2 * 0 = 0 or x = 2 * 1 = 2 or x = 2 * 2 = 4
N={0,2,4}
M∩N＝{0,2}

Let m = {a, 2}, n = {X / X-1 △ x + 1 ≤ 0, X ∈ Z], if M ∩ n = {0], Mun = P, then there are A4, B8, C16 and D32 subsets of P

Please write the set n expression under the specification

Given the set M = {X / x ^ 2-x > 0}, and the set n = {X / x > = 1}, then Mun=

M={x|x²-x>0}={x|x(x-1)>0}={x|x1}
MUN={x|x

Solving the equation x / (x + 2a) - X / (x-2a) = A & # 178; / (4a & # 178; - X & # 178;) (a ≠ 0)

X (x-2a) - x (x + 2a) = - A & # 178;
The results show that: - 4ax = - A & # 178;
The solution is: x = A / 4

If x = 1 is the solution of the equation A & # 178; X + 2A = 2x, then - 4A + 5-2a & # 178=___

a²x+2a=2x
a²+2a=2
-a²-2a=-2
-2a²-4a=-4
-4+5=1

If the equation (4 / X & # 178; - 6x + 4a) + (1 / x-2a) = 0 has a root of 6, try to find the value of A

Because 6 is the solution of the equation, the value of a can be obtained by substituting 6 into the equation and simplifying the solution

Solving inequality 4A & # 178; - 16a-4 > 0

4a²-16a-4>0
4(a²-4a+4-5)>0
a²-4a+4-5>0
(a-2)²>5
A-2 > √ 5 or A-22 + √ 5 or a