It is known that the solution of the equation 2x-a / 11-A = X-2 / 2 + A / 3 about X is - 2, and the value of a is obtained

It is known that the solution of the equation 2x-a / 11-A = X-2 / 2 + A / 3 about X is - 2, and the value of a is obtained

The original equation is simplified
x-47/33a+1=0
a=33/47*(x+1)
If we take - 2 into the original equation, we get
a=-33/47

Given that x = 5 is the solution of the equation 2x + a-11 = 0 about X, then the value of a is

Because x = 5 is the solution of the equation, so 2 * 5 + a-11 = 0, so a = 1

Given that the line L1 passes through point a (- 1,0) with a slope of K, the line L2 passes through point B (1,0) with a slope of - 2 / K, where k is not equal to 0, and the line L1 and L2 intersect at point m, (1) find the trajectory equation of the moving point m, (2) if the trajectory of the line L passing through point n (1 / 2,1) intersects point m at two points c and D, and N is the midpoint of line CD, find the equation of the line L

Line L1: y = KX + k
Line L2: y = - 2 / k * x + 2 / K
The multiplication of both sides of the two formulas yields:
x2+y2=1
Remove the four points that intersect the two axes
That is the trajectory equation of M
(2) Let L: y = KX + 1-k / 2
Substituting x2 + y2 = 1, we get the following result:
（1+k2)x2+2kx-3k2/4-k=0
x1+x2=-k/(1+k2)
And X1 + x2 = 1
So - K / (1 + K2) = 1
There is no solution for K
N cannot be the midpoint of the segment CD

we don't()()feed the goldfish too much food.

need to
We don't need to feed goldfish too much

Given that the parabola y = ax ^ 2-2x + 1 has no intersection with the X axis, is the quadrant where the vertex of the parabola lies in the fourth quadrant

No
∵ its intersection with Y-axis is (0,1), and it has no intersection with x-axis
The opening is upward and the vertex is above the x-axis,
The quadrant of the vertex is not in the fourth quadrant
2 & # 178; - 4A1, axis of symmetry x = - (- 2) / (2a) = 1 / A

When a natural number greater than 1 is removed from 300243205 and the same remainder is obtained, then the natural number is______ ．

Let this natural number be x, a, B and C be natural numbers, and a be a constant, because when this number is divided by 300243205, we get the same remainder, so ax + a = 300 (1); BX + a = 243 (2); CX + a = 205 (3); (1) - (2) get, (a-b) x = 57, (2) - (3) get, (B-C) x = 38, (1) - (3)

If a hyperbola passes through a point (6, √ 6) and its asymptote equation is y = ± 1 / 3x, then the hyperbolic equation is?

Let X & sup2; / A & sup2; - Y & sup2; / B & sup2; = ± 1
Then ± B / a = ± 1 / 3, B / a = 1 / 3, a = 3B
Substituting (6, √ 6) into X & sup2; / (9b & sup2;) - Y & sup2; / B & sup2; = 1 and X & sup2; / (9b & sup2;) - Y & sup2; / B & sup2; = - 1 respectively
We obtain that B & sup2; has unique solution 2, and the hyperbolic equation is X & sup2; / A & sup2; - Y & sup2; / B & sup2; = - 1
So a & sup2; = 9b & sup2; = 18
Hyperbola X & sup2; / 18-y & sup2; / 2 = - 1
Simplification
Hyperbola Y & sup2; / 2-x & sup2; / 18 = 1

On "integral multiplication and division and factorization" in mathematics of grade two
Factorization
4x²-40xy+25y²

4x²-40xy+25y²
=4（x²-10xy+25y²）-75y²
=4（x-5y)²-(5√3y)²
=(2x-10y)²-(5√3y)²
=(2x-10y+5√3y)(2x-10y-5√3y)

Given the image of quadratic function y = ax square - 4x + C, the analytic expression of the quadratic function is obtained through a (- 1,0) B (4, - 5)

The image of y = ax square - 4x + C passes through a (- 1,0) B (4, - 5)
=》
0=a+4+c
-5=16a-16+c
The solution is a = 1, C = - 5
The analytic formula of this function is y = xsquare-4x-5

Given the square of the parabola y = a (X-H), when x = 2, there is a maximum value. This parabola passes through (1, - 3), find the analytical formula of the parabola, and point out that when x is the value, y decreases with the increase of X?

y=a(x-h)^2,x=2,【1,-3】
Y = ax ^ 2-4ax + 4A
If a-4a + 4A = - 3, then a = - 3
Y = - 3x ^ 2 x 12x-12