Let m = {x | - 1 ≤ XA}, and m ∩ n ≠ & # 8709;, the value range of A

Let m = {x | - 1 ≤ XA}, and m ∩ n ≠ & # 8709;, the value range of A

M = {x | - 1 ≤ XA}, and m ∩ n ≠ &;
Then a < 2
If you don't understand, please hi me, I wish you a happy study!

Let m = {x | 0 ＜ x ＜ 2}, n = {x | x ＞ a}, if M ∩ n = & # 8709;, then the value range of a is the solution

The solution is m = {x | 0 ＜ x ＜ 2}, n = {x | x ＞ a}, if M ∩ n = & # 8709;,
Drawing on the number axis shows that
That is, a ≥ 2

Let a = {x | x square-x-2 = 0}, B = {x | x-m | 1, x, m ∈ r}, if a ∩ B = & # 8709;, find the value range of M

x²-x-2=0
(x+1)(x-2)=0
We get x = - 1, x = 2
So a = {- 1,2}
|x-m|＜1
-1＜x-m＜1
m-1＜x＜m+1
Because a ∩ B = &;,
therefore
M-1 ＞ - 1, m ＞ 0
M + 1 < 2, m < 1
That is, the value range of M is 0 < m < 1

It is known that the set a = {x belongs to R | x + 2|

Set a = {x ∈ R | x + 2|

Given the set M = {(x, y) | x + y = 1}, n = [(x, y) | X-Y = 1}, then m ∩ n is equal to ()
A. x=1，y=0B. （1，0）C. {（1，0）}D. {1，0}

∵ M = {(x, y) | x + y = 1}, n = [(x, y) | X-Y = 1}, ∩ m ∩ n = {(x, y) | x + y = 1x − y = 1 = {(1, 0)} so select C

Given the set M = {(x, y) | Y & sup2; = 2x,}, P = {(x, y) | x + y + a = 0}, then m ∩ P = an empty set,

Set M = {(x, y) | Y & # 178; = 2x,}, P = {(x, y) | x + y + a = 0}, then m ∩ P = empty set,
Then y & # 178; = 2x and X + y + a = 0 have no solution,
Eliminate x
y²/2+y+a=0,
Then △ = 1-4 * 1 / 2 * a = 1-2a < 0,
a＞1/2.
That is, m ∩ P = empty set if and only if a > 1 / 2

The equation x & sup2; + Y & sup2; + 2aX + by + C = 0 indicates that the center of the circle is (2,2), the radius of the circle is 2, then ABC is?
Sorry, the symbol in front of "by" is "- by, and your answers are not in the range of choice

a=-2 b=4 c=4
In this way, you can write the equation for a circle with center (2,2) and radius 2: x2-4x + y2-4y + 4 = 0, and then compare it with the original equation

In high school mathematics, it is known that a straight line passes through a point (- 1,3) and is tangent to a circle C: x ^ 2 + y ^ 2-2x-3 = 0, so the equation of a straight line can be solved

Let the linear equation be y = K (x + 1) + 3 circle C: x ^ 2 + y ^ 2-2x-3 = (x-1) ^ 2 + y ^ 2-4 = 0 (x-1) ^ 2 + y ^ 2 = 4, the center of the circle is (1,0), the radius is 2, so the distance from (1,0) to the tangent d = 2|2k + 3 | / √ (1 + K ^ 2) = 2 (2k + 3) ^ 2 = 4 (1 + K ^ 2) 12K + 5 = 0k = - 5 / 12 linear equation: y = - 5 / 12 * (x + 1) + 3, that is: 5x + 12y-31 =

Given m ∈ R, a ＞ B ＞ 1, f (x) = mxx − 1, try to compare the size of F (a) and f (b)

When f (a) - f (b) = MAA − 1-mbb − 1 = m (B − a) (a − 1) (B − 1) ∵ a ＞ B ＞ 1, ∵ A-1 ＞ 0, B-1 ＞ 0, B-A ＜ 0, ∵ m ＞ 0, m (B − a) (a − 1) (B − 1) ＜ 0, ∵ f (a) ＜ f (b); when m = 0, m (B − a) (a − 1) (B − 1) = 0, ∵ f (a) = f (b); when m ＜ 0

Given m ∈ R, a ＞ B ＞ 1, f (x) = mxx − 1, try to compare the size of F (a) and f (b)

This is where f (a) - f (a) - f (b) (f (a) - f (b) (f (a) - f (b) (f (a) - f (a) - f (a) - f (b) (B-B) (B-B) (B-B − 1 = m (B − a) (A-1) (A-1) (a-mbb − 1 = m (b) (b) (B-B \ (a) - B (b) (B-B − a) (a (a) (a) (a) (a) (a) (a) (a) (a) (a (a) (b) (b) (b) (b) (a) (a) (a) (b) (b) (b) (b) (b) (b) (b) (b) (b) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b(a) > F (b)