Given that two circles C1: (x + 4) 2 + y2 = 2, C2: (x-4) 2 + y2 = 2, the moving circle m is tangent to both circles C1 and C2, then the trajectory equation of the moving circle center m is () A. X = 0b. X22-y214 = 1 (x ≥ 2) C. x22-y214 = 1D. X22-y214 = 1 or x = 0

Given that two circles C1: (x + 4) 2 + y2 = 2, C2: (x-4) 2 + y2 = 2, the moving circle m is tangent to both circles C1 and C2, then the trajectory equation of the moving circle center m is () A. X = 0b. X22-y214 = 1 (x ≥ 2) C. x22-y214 = 1D. X22-y214 = 1 or x = 0

If two definite circles and moving circles are circumscribed or both inscribed, that is, two circles C1: (x + 4) 2 + y2 = 2, C2: (x-4) 2 + y2 = 2, moving circle m is tangent to both circles C1 and C2, that is, point m is on the vertical bisector of line C1 and C2, and the coordinates of C1 and C2 are (- 4,0) and (4,0) respectively. Its vertical bisector is y-axis, and the trajectory equation of moving circle center m is x = 0 A circumscribe, let it be inscribed with circle C1: (x + 4) 2 + y2 = 2, and circumscribed with circle C2: (x-4) 2 + y2 = 2, then the distance from m to (4,0) is reduced to (- 4,0), and the difference is 22. According to the definition of hyperbola, the locus of point m is a hyperbola with (- 4,0) and (4,0) as the focus and 2 as the length of real half axis, so B2 = c2-a2 = 14, so the equation of hyperbola is x22-y214 = 1 It is known that the trajectory equation of moving circle m is x22-y214 = 1 or x = 0, D
The moving circle C and the fixed circle C1: x ^ 2 + (y-4) ^ 2 = 64 are inscribed, and the fixed circle C2: x ^ 2 + (y + 4) ^ 2 = 4 are circumscribed
X ^ 2 + (y-4) ^ 2 = 64 circle center a (0,4) radius 8x ^ 2 + (y + 4) ^ 2 = 4 circle center B (0, - 4) radius 2 let the moving circle center 0 'the radius of the moving circle r from the title 8-r = | o'a | 2 + r = | o'b | to get | o'a | + | o'b | = 10, that is, the trajectory of o' is an ellipse with the distance sum of 10 from two focal points a and B. let x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, then 2A = 10 a = 5C
Ellipse with focus on X-axis a = 5, B = 3, C = 4
After drawing, you will find that the track is ellipse and has a range, and the focus should be on the Y axis.
Given that the solution of equation 3 (x-2a) + 2 = x-a + 1 is suitable for the inequality X-5 / 2 ≥ 4a, the value range of a is obtained
a≤-1
From the equation: x = 5a-1 / 2
Substituting the inequality, a ≤ - 1
There may be miscalculation, the landlord or check it!
Firstly, the solution of equation 3 (x-2a) + 2 = x-a + 1 is obtained, and x = (5a-1) / 2 is brought into the inequality to remove X-5 / 2 ≥ 4a
It is obtained that (5a-1) / 2-5 / 2 ≥ 4A solves the inequality a ≤ - 2
From the equation, x = 5 / 2A + 1 / 2
The inequality leads to a ≤ - 2
If the solution set of the inequality system 2a-x > 3,2x + 8 > 4A about X does not have the value of every x in - 1
2a-x>3 x4a x>2a-4
2a-4
2a-3>x,2x>4a-8,x>2a-4,2a-4
It is known that the solution of equation 3 (x-2a) = x-a + 1 is suitable for the inequality x-3 ≤ 4a, and the value range of a is obtained
∵ 3 (x-2a) = x-a + 1, ∵ 3x-6a = x-a + 1, sorted out as 2x = 5A + 1, ∵ x = 5A + 12; ∵ the solution of the equation 3 (x-2a) = x-a + 1 about X is suitable for the inequality x-3 ≤ 4a, ∵ 5A + 12 − 3 ≤ 4a, the denominator is removed to get 5A + 1-6 ≤ 8a, ∵ 3a ≥ - 5, a ≥ − 53, ∵ the value range of a ≥ − 53
If the inequality system {X-1 > = 2A and X + 9 ＜ 4A about X have no solution, the value range of a is obtained
x-1>2a x>2a+1
X+9
If the solution set of x-2a + B0 is - 1
x-2a+b0
Item transfer: 2x > 5b-3a
Divide both sides by two at the same time
x＞5b-3a/2
Because the solution set of the inequality system x-2a + B0 is - 1
x-2a+b0
Item transfer: 2x > 5b-3a
Divide both sides by two at the same time
x＞5b-3a/2
Because the solution set of the inequality system x-2a + B0 is - 1
Given that the solution set of inequality system x-2a + B < 02x + 3a-5b > 0 is - 1 < x < 6, then a=______ ，b=______ .
By simplifying the inequality system, the solution set of X ＜ 2a-bx ＞ 5b-3a2 ∵ is - 1 ＜ x ＜ 6 ∵ 2a-b = 6, 5b-3a2 = - 1, and the solution set of a = 4, B = 2
The following inequality must be true is () a.4a > 3A B. - a > - 2A c.3-x2/a
The following inequality must be (c) a.4a > 3A B. - a > - 2A c.3-x2/a
A. B, d three answers, as long as the a = - 1 into can be considered not true
If the solution set of the inequality system {2x + 3a-5b ＞ 0 is - 1 ＜ x ＜ 6, the values of a and B. {x-2a + B ＜ 0
From: 2x + 3a-5b > 0
x-2a+b＜0
The results show that: (5b-3a) / 2 ＜ x ＜ 2a-b
Because: the solution set of X is - 1 < x < 6
So: (5b-3a) / 2 = - 1
2a-b=6
The solutions of the equations are: a = 4, B = 2