x. Y are all natural numbers, and X (X-Y) - Y (Y-X) = 12, find the value of X + y-xy

x. Y are all natural numbers, and X (X-Y) - Y (Y-X) = 12, find the value of X + y-xy

x(x-y)-y(y-x)=12
Then: (x + y) * (X-Y) = 12
x. Y is a natural number, so X1 = 2, Y1 = 4
x2=4,y2=2
So x + y-xy = - 2

10. Y is a natural number, X (X-Y) - Y (Y-X) = 12, find the value of x.y

(x+y)(x-y)=12
Because XY is a natural number
So x = 4, y = 2

Factorization 1 + X + x ^ 2 + x ^ 3 + '+ x ^ 15

1+x+x^2+x^3+… +x^15
=(1+x)+(x^2+x^3)+(x^4+x^5)… +(x^14+x^15)
=(1+x)+x^2(1+x)﹢x^4(1+x)﹢… ﹢x^14(1+x)
=(1+x)(x^2+x^4+… +X ^ 14) (same as below)
=(1+x)(1+x²)(1+x^4+x^8+x^12))
=(1+x)(1+x²)(1+x^4)(1+x^8)

There are several numbers, of which the first one is - 1 / 3. From the second one, every number after is equal to the reciprocal of the difference between 1 and the previous one
(1) Calculate the value of the second, third and fourth number respectively
(2) Calculate the sum of the first 36 numbers

(1) the second 1 △ 1 - (- 1 / 3)) = 3 / 4,
The third 1 △ 1 - (3 / 4)) = 4,
The fourth one (1-4) = - 1 / 3
(2) the sum of the first 36 numbers
=﹙-1/3)+3/4+4＋… ＋﹙-1/3)+3/4+4
＝[﹙-1/3)+3/4+4]×12
= -4+9+48
=53

9. The trajectory equation of the point whose distance to the line L: 3x + 4y-5 = 0 is equal to 1 is ()
A.3x+4y-4=0
B. 3x + 4Y = 0 or 3x + 4y-10 = 0
C.3x+4y+10=0
D. 3x + 4y-30 = 0 or 3x + 4Y + 20 = 0
Mainly why
Then the calculated point (2,1)?

First of all, we can be sure that there are two tracks of a point, one on one side, so we can only choose B or D
Two straight lines in B, one goes through (0,0) and the other goes through (2,1). The distance between these two points and the known straight line is equal to 1, so B is correct
The second line, you can bring (2,1) into the linear equation, you can know whether it passed this point!

Let f (x) defined on R satisfy f (x) · f (x + 2) = 13, if f (1) = 2, then f (99) = ()
A. 13B. 2C. 132D. 213

∵ f (x) · f (x + 2) = 13 and f (1) = 2 ∵ f (3) = 13F (1) = 132, f (5) = 13F (3) = 2, f (7) = 13F (5) = 132, f (9) = 13F (7) = 2, ∵ f (2n − 1) = 2 & nbsp; n is odd 132 & nbsp; n is even, f (99) = f (2 × 100 − 1) = 132, so C is selected

Using formula method to solve equation 1 / 3x & sup2; - 2X-4 = 0
Please write down the calculation process,

Formula method for solving equation 1 / 3x & sup2; - 2X-4 = 0
First, we get X & sup2; - 6x-12 = 0
a=1 b=-6 c=-12
X = - B ± √ b square - 4ac / 2A
X=6±√6×6+4×1×12/2×1
X=3±√21
X1=3+√21 X2=3-√21

Intersection and symmetry axis of quadratic function image and X axis
The intersection of the image of the function y = (x + 1) (X-2) and the X axis is_____ The axis of symmetry is______

The point of intersection with x-axis is the solution of equation = 0
therefore
（x=-1 x=2)
(-1 0)(2 0)
y=x²+x-2x-2
=x²-x-2
Axis of symmetry X-B / 2A = 1 / 2
So the axis of symmetry x = 1 / 2

*(9 + a) - * (16-3a) + * (4a + 1) + * (- A & sup2;) [A is a real number]
*Meaning: under the root

Starting from (- A ^ 2) under the root sign,
If - A ^ 2 ≥ 0, then a ^ 2 ≤ 0 ∥ a = 0
In this way, the original formula becomes
Root 9 - root 16 + root 1 + root 0
=0

If the equation (3-m) x ^ | m | - 5 + 2 = 3 is a linear equation with one variable, then M=

(3-m) x ^| m | - 5 + 2 = 3 is a linear equation with one variable
The degree of X is equal to 1 and the coefficient is not equal to 0
|M | - 5 = 1, 3-m ≠ 0, M = ± 6