Given that the distance between two points a and B is 2, find the trajectory equation of the point whose square difference of the distance between two points is 1

Given that the distance between two points a and B is 2, find the trajectory equation of the point whose square difference of the distance between two points is 1

Let a be (1,0), B be (- 1,0), and the points that meet the requirements are (x, y)
(x-1) ^ 2 + y ^ 2 - [(x + 1) ^ 2 + y ^ 2] = soil 1
-4X = soil 1
X = - 1 / 4 or x = 1 / 4
So the trajectory equation of the point is a straight line whose distance between two points is 3 / 4 and 5 / 4 respectively

Given that the ratio of distance between point m and two fixed points o (0,0) and a (3,0) is 12, the trajectory equation of point m is ()
A. x2+y2+2x-5=0B. x2+y2+2x-3=0C. x2+y2-2x-5=0D. x2+y2-2x-3=0

Let m (x, y), the ratio of the distance between point m and two fixed points o (0, 0), a (3, 0) be 12, then we get x2 + Y2 (x − 3) 2 + y2 = 12, then we get: x2 + Y2 + 2x-3 = 0. The trajectory equation of point m is x2 + Y2 + 2x-3 = 0

Given that the ratio of distance between point m and two fixed points o (0.0) a (3.0) is 1:2, the trajectory equation of point m can be obtained

Let m (x, y)
MO/MA=1/2
That is mo ^ 2 / MA ^ 2 = 1 / 4
4*MO^2=MA^2
The results are as follows
4(x^2+y^2)=(x-3)^2+y^2
The result is: 3x ^ 2 + 3Y ^ 2 + 6x-9 = 0
Namely: x ^ 2 + y ^ 2 + 2x-3 = 0

Given that the ratio of distance between point m and two fixed points o (0,0) and a (3,0) is 12, the trajectory equation of point m is ()
A. x2+y2+2x-5=0B. x2+y2+2x-3=0C. x2+y2-2x-5=0D. x2+y2-2x-3=0

Let m (x, y), the ratio of the distance between point m and two fixed points o (0, 0), a (3, 0) be 12, then we get x2 + Y2 (x − 3) 2 + y2 = 12, then we get: x2 + Y2 + 2x-3 = 0. The trajectory equation of point m is x2 + Y2 + 2x-3 = 0

Given (x-2y) ^ + | 5x-7y-21 | = 0, find the value of X, y
mathematics

From the original problem, we can see that x-2y = 0, 1, 5x-7y-21 = 0, 2. From 1, we get x = 2Y, bring in 2, we get 3Y = 21, y = 7. Therefore, x = 14, y = 7

Given that the sum of solutions X and y of the equation system 5x − 7Y = 5m + 42x − y = 3 − m is not less than 4, then the value range of M is______ ．

If 5x − 7Y = 5m + 4, ① 2x − y = 3 − m, ② - ② × 7, the sum of solutions X and y of the equation system 5x − 7Y = 5m + 42x − y = 3 − m is not less than 4, ｜ 17 − 12m9 + 7 − 15m9 ≥ 4, m ≤ - 49

It is known that the sum of X, y of the equation system 5x-7y = 5m + 4 2x-y = 3-m is not less than the range of 4, M
The one who knows taught me that the best way is to get the solution down. The child asked me, but I forgot

5x-7y=5m+4 (1)
2x-y=3-m (2)
From (2), y = 2x-3 + m (3)
(3) Substituting (1) gives x = (17-12m) / 9 (4)
(4) Substituting (3) gives y = (7-15m) / 9
∴x+y=(8-9m)/3>=4
m

For polynomial x ^ 4 + 4, factorize it

=x^4+4x^2+4-4x^2
=(x^2+2)^2-4x^2
=(x^2+2x+2)(x^2-2x+2)

If for any allowable value in a certain range, P = | 1-2x | + | 1-3x | + +|If the value of 1-9x | + | 1-10x | is constant, then the value is ()
A. 2B. 3C. 4D. 5

Since 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10, the value range of X is: 1-7x ≥ 0 and 1-8x ≤ 0, that is, 18 ≤ x ≤ 17, so p = (1-2x) + (1-3x) + +(1-7x) - (1-8x) - (1-9x) - (1-10x) = 6-3 = 3

If the value of 2x + | 4-5 | + | 1-3x | + 4 is constant, what conditions should x satisfy and what is the value of this constant|
Which do it for me? I'll offer a reward of 50 points

2X+/4-5/+/1-3X/+4=2X+/1-3X/+3
If 1-3x > 0, then X