1 / 12 = (1 / x) + (1 / y) + (1 / z), XYZ are three different natural numbers. What are the three numbers?

1 / 12 = (1 / x) + (1 / y) + (1 / z), XYZ are three different natural numbers. What are the three numbers?

There are many solutions
such as
1/12=2/24=1/24+1/24=1/24+3/72=1/24+2/72+1/72=1/24+1/36+1/72
So 24, 36, 72
1/12=3/36=1/36+2/36=1/36+3/54=1/36+2/54+1/54=1/36+1/27+1/54
27, 36, 54
……

X / y = 5 (x, y is a non-zero natural number), the greatest common factor of X and Y is (), and the least common multiple is ()

The greatest common divisor is y and the least common multiple is X

X and y are natural numbers, X △ y = 3 (Y ≠ 0), and the greatest common divisor of X and Y is______ The least common multiple is______ ．

X and y are both natural numbers, X △ y = 3 (Y ≠ 0), that is, X and y are multiple relations, then the greatest common divisor of X and Y is y, and the least common multiple is x; so the answer is: y, X

Given x + y + Z + XY + XZ + YZ + XYZ = 182 (where x, y, Z are all natural numbers, and x > y > z), find the value of X, y, Z

(x + 1) (y + 1) (Z + 1) = 182
Because 182 = 2 * 7 * 13, x = 12, y = 6, z = 1

It is known that x + y + Z, XY + YZ + ZX and XYZ are integers. It is proved that x ^ n + y ^ n + Z ^ n is an integer (n is any natural number)

According to Weida's theorem, x, y, Z are the three roots of the equation T ^ 3 - (x + y + Z) T ^ 2 + (XY + YZ + XZ) t-xyz = 0
Take X and multiply both sides by x ^ n to get x ^ (n + 3) - (x + y + Z) x ^ (n + 2) + (XY + YZ + ZX) x ^ (n + 1) - xyzx ^ n = 0
For y and Z, we can get the same formula. By adding the three formulas and merging the same term, we can recursively prove that x ^ n + y ^ n + Z ^ n is an integer by knowing that x + y + Z, XY + YZ + ZX and XYZ are integers

Given x ^ 2n + 1 * x ^ 4 = x ^ n-2 * x ^ 3n-1, then the value of n is ()

x^(2n+1）*x^4=x^(n-2)*x^(3n-1)
2n+1+4=n-2+3n-1
2n-4n=-5-2-1=-8
n=4

Given X2N = 4, find the value of (x3n) 2-xn (where x is a positive number and N is a positive integer)

∵ X2N = 4, X is a positive number, n is a positive integer, ∵ xn = 2, ∵ (x3n) 2-xn = (xn) 6-xn = 26-2 = 62

Given that n is a positive integer and x ^ 2n = 4, find 3 (x ^ 3n) ^ 2-13 (x ^ 2) ^ 2n,

Original title = 3 (x ^ 2n) ^ 3-13 (x ^ 2n) ^ 2 = 2 * 4 ^ 3-13 * 4 * 2 = 24

Given x ^ (2n) = 5, find the value of [1 / 3x ^ (3n)] ^ (2) - 3 * [x ^ (2)] ^ (2n)

Solution
simple form
=1/9[x^(2n)]³-3×[x^(2n)]²
=1/9×5³-3×5²
=125/9-75
=550/75
=22/3

If 2n of x = 2. Find the square of (2x + y) / (- 2x-y) to the fourth power of

(3N power of - x) to the second power of - 4 (- square of x)
=6N power of x-2n power of 4x
=(x has 2n power) & the 2n power of # - 4 × x
=8-8
=0
Cubic power of (2x + y) * (2x + y) square / (- 2x-y) quartic power
=The 5th power of (2x + y) / the 4th power of (2x + y)
=2x+y