A (- 3,2), B (3, - 2), prove that the trajectory equation of the point m whose distance is equal to the two points is 3x-2y = 0

A (- 3,2), B (3, - 2), prove that the trajectory equation of the point m whose distance is equal to the two points is 3x-2y = 0

Let the coordinates of point m be (x, y)
|MA|=|MB|
[(-3-x)^2+(2-y)^2]^(1/2)=[(3-x)^2+(-2-y)^2]^(1/2)
[(-3-x)^2+(2-y)^2]=[(3-x)^2+(-2-y)^2]
9+6x+x^2+4-4y+y^2=9-6x+x^2+4+4y+y^2
12x-8y=0
3x-2y=0
That is to say
(1) What are the quadratic and primary terms in the polynomial 3x & # 178; + 2x + 9?
(2) Given that a and B are constant terms, the difference between ax & # 178; + 3xy-5x and 2x & # 178; - 2bxy + 2Y does not contain a quadratic term, the value of B to the power of a-5 / 4 can be obtained
Please!
The second term is 3x & #, and the first term is 2x
Difference = (A-2) x & # 178; + (3 + 2b) xy-5x-2y does not contain quadratic term
Then A-2 = 0
3+2b=0
So a = - 2, B = - 3 / 2
So the original formula = 9 / 4-5 / 4 = 1
(1) The quadratic term in the polynomial 3x & # 178; + 2x + 9 is (3x & # 178;), and the primary term is (2x)
(2) Given that a and B are constant terms, the difference between ax & # 178; + 3xy-5x and 2x & # 178; - 2bxy + 2Y does not contain a quadratic term, the value of a-power-5 / 4 of B is obtained.
ax²+3xy-5x-(2x²-2bxy+2y)
=(a-2)x²+(3+2b)xy-5x-2y
Because there's no expansion in the result
(1) The quadratic term in the polynomial 3x & # 178; + 2x + 9 is (3x & # 178;), and the primary term is (2x)
(2) Given that a and B are constant terms, the difference between ax & # 178; + 3xy-5x and 2x & # 178; - 2bxy + 2Y does not contain a quadratic term, the value of a-power-5 / 4 of B is obtained.
ax²+3xy-5x-(2x²-2bxy+2y)
=(a-2)x²+(3+2b)xy-5x-2y
Because there is no quadratic term in the result, then
a-2=0,a=2
3+2b=0,b=-3/2
So, B ^ a-5 / 4 = (- 3 / 2) ^ 2-5 / 4 = 9 / 4-5 / 4 = 1
(1) The second term is 3 and the first term is 2
(2) (AX ^ 2 + 3xy-5x) - (2x ^ 2-2bxy + 2Y) = (A-2) x ^ 2 + (3 + 2b) xy-5x-2y (excluding quadratic term, coefficient is 0)
Because A-2 = 0
3+2b=0
So a = 2, B = - 3 / 2
The final solution is obtained
The solution {x-2y = 0 {5x-7y-2 = 0}
x=4/3
y=2/3
Substituting x = 2Y into 5x-7y-2 = 3y-2 = 0
So y = 2 / 3
X=4/3,Y=2/3
Given the set a = {- 3x, 2x-1, - 4}, B = {x-5,1-x, 9} and a intersects B, find the value of X
Suppose: - 3x = 9, then x = - 3, substituting a / B,
Suppose, - 3x = X-5, then x = 1.2 is substituted into a / b
Suppose, - 3x = 1-x, then x = - 0.5 is substituted into a / b
Suppose, - 2x = 9, then x = - 4.5 is substituted into a / b
Suppose that 2x-1 = X-5, then x = - 4 is substituted into a / b
Suppose that 2x-1 = 1-x, then x = 2 / 3 is substituted into a / b
What is a given to B? You may have copied the title wrong. Go back and check it.
From the meaning of the question, we can get a ∩ B ≠ &
(1) The solution of - 3x = X-5 is x = 5 / 4
(2) The solution of - 3x = 1-x is x = - 1 / 2
(3) The solution of 2x-1 = X-5 is x = - 4
(4) The solution of 2x-1 = 1-x is x = 2 / 3
(5) The solution of - 3x = 9 is x = - 3
(6) 2x-1 = 9 and x = 5
(7) X-5 = - 4 gives x = 1
(8) 1-x = - 4, the solution is x = 5
Equations {4x-7y = 5,3x + 2Y = 12
4x-7y=5 (1)
3x+2y=12 (2)
(1)×2+(2)×7
8x+21x=10+84
29x=94
therefore
x=94/29
y=(4x-5)/7=33/29
4x-7y=5。。。。。 One
3x+2y=12.。。。。 Two
The formula 1 is multiplied by 3 to get 12x-21y = 15.... Three
Formula 2 times 4 gives 12x + 8y = 48.... Four
Formula 3 and 4 are subtracted
29y=33
Then y = 33 / 29
Then x = 94 / 29
Let a = {x ^ 2,2x-1, - 4}, B = {x-5,1-x, 9}, if a intersects B = {9}, find a and B?
Because a intersects B = {9},
therefore
9 belongs to a
When x ^ 2 = 9
x=3,-3
X = 3, a = {9,5, - 4}, B = {- 2, - 2,9}
x=-3,A={9,-7,-4},B={-8,4,9}
Then a and B = {9, - 7, - 4, - 8,4}
When 2x-1 = 9, x = 5
A={25,9,-4},B={0,-4,9},
It doesn't satisfy a intersection B = {9}, so it is discarded
So a and B = {9, - 7, - 4, - 8,4}
Because a ∩ B = {9}
So there is 9 in a, so 9 = x & sup2; or 9 = 2x-1
We get x = - 3, x = 3 or x = 5
If x = - 3, then a = {9, - 7, - 4}, B = {- 8,3,9}
If x = 3, then a = {9,5,4}, B = {- 2, - 2,9}, there are repeated elements in B, which are not consistent with each other
If x = 5, then a = {25,9, - 4}, B = {0, - 4,9}, a ∩ B = {- 4,9}, does not satisfy the condition... Expansion
Because a ∩ B = {9}
So there is 9 in a, so 9 = x & sup2; or 9 = 2x-1
We get x = - 3, x = 3 or x = 5
If x = - 3, then a = {9, - 7, - 4}, B = {- 8,3,9}
If x = 3, then a = {9,5,4}, B = {- 2, - 2,9}, there are repeated elements in B, which are not consistent with each other
If x = 5, then a = {25,9, - 4}, B = {0, - 4,9}, a ∩ B = {- 4,9}, does not meet the conditions, and is discarded
So a = {9, - 7, - 4}, B = {- 8,3,9}, a ∪ B = {9, - 7, - 4, - 8,3} put away
A intersects B = {9}, so x ^ 2 = 9 or 2x-1 = 9, that is, x = - 3 or 3 or 5.
1. When x = - 3, a = {9, - 7, - 4}, B = {- 8, 4, 9} meets the condition.
2. When x = 3, a = {9,5, - 4}, there are two elements in B equal to - 2, so x cannot be equal to 3
3. When x = 5, a = {25,9, - 4}, B = {0, - 4,9}, then a intersects B = {- 4,9}, which does not meet the condition.
To sum up, x = - 3. So a and B = {- 8, - 7... Expand
A intersects B = {9}, so x ^ 2 = 9 or 2x-1 = 9, that is, x = - 3 or 3 or 5.
1. When x = - 3, a = {9, - 7, - 4}, B = {- 8, 4, 9} meets the condition.
2. When x = 3, a = {9,5, - 4}, there are two elements in B equal to - 2, so x cannot be equal to 3
3. When x = 5, a = {25,9, - 4}, B = {0, - 4,9}, then a intersects B = {- 4,9}, which does not meet the condition.
To sum up, x = - 3. So a and B = {- 8, - 7, - 4, 4, 9} put away
Given equation 3 (X-Y) - 5x + 12 = 2x-7y-4, then the value of X-Y is_
3(x-y)-5x+12=2x-7y-4
3(x-y)+12=5x+2x-7y-4
3(x-y)+12=7(x-y)-4
4(x-y)=16
x-y=4
Four
Given the equation 3 (X-Y) - 5x + 12 = 2x-7y-4, then the value of X-Y is u 4
4x^2[2x-(x^2-2x+3)-2(x^2-3x)]
4x^2[2x-(x^2-2x+3)-2(x^2-3x)]
=4x²(2x-x²+2x-3-2x²+6x)
=4x²(-3x²+6x-3)
=-12x^4+24x^3-12x^2;
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Given equation 3 (X-Y) ^ 2-5x-2 = 2x-7y-4. Find the value of X-Y
The equation is: 3 (X-Y) ^ 2-7 (X-Y) + 2 = 0
That is, (x-y-2) [3 (X-Y) - 1] = 0
So X-Y = 2 or X-Y = 1 / 3
Let X-Y = Z
Then the equation becomes 3Z ^ 2-7z + 2 = 0
Z = 2 or - 1 / 3
X-Y = 2 or - 1 / 3
3(x-y)^2-2=5x-5y+2x-2y-4
3(x-y)^2-2=5(x-y)+2(x-y)-4
3(x-y)^2-7(x-y)+2=0
X-Y = 1 / 3 or 2
4(3x^2-x-1)(x^2+2x-3)-(4x^2+x-4)^2
4(3x^2-x-1)(x^2+2x-3)-(4x^2+x-4)^2=4(3x^2-x-1)(x^2+2x-3)-[(3x^2-x-1)+(x^2+2x-3)]^2=4(3x^2-x-1)(x^2+2x-3)-(3x^2-x-1)^2-2(3x^2-x-1)(x^2+2x-3)-(x^2+2x-3)^2=-(3x^2-x-1)^2+2(3x^2-x-1)(x^2+2x-3)-(x^2+2x-3)^...