Let x ^ 2n = 3, find (1 / 3x ^ 3n) · [4 (x ^ 5) ^ n] Let x ^ 2n = 3, find (1 / 3x ^ 3n) ^ 3 · [4 (x ^ 5) ^ n]

Let x ^ 2n = 3, find (1 / 3x ^ 3n) · [4 (x ^ 5) ^ n] Let x ^ 2n = 3, find (1 / 3x ^ 3n) ^ 3 · [4 (x ^ 5) ^ n]

(1/3x^3n)·[4(x^5)^n]
=1/27*x^3n*4*x^5n
=(4/27)*x^8n
=(4/27)*(x^2n)^4
=(4/27)*3^4
=(4/27)*81
=4*3
=12

If A2N + 1b2 and 5a3n-2b2 are of the same kind, then n=______ ．

According to the definition of similar term, 2n + 1 = 3n-2, the solution is n = 3

One kind of operation h is 1, when n is odd, H = 3N + 13 2, when n is even, H = n * 1 / 2 * 1 / 2 * 1 / 2... (H is odd) seek 257 times of H operation results
Another question: how many groups of integer solutions of the equation 'xy-3x-5 and the sum of the absolute values of Y is 0'?
Double reward

1.257 = 784 = 49 = 160 = 5 = 28 = 7 = 34 = 17 = 64 = 1 = 16 = 1 = start cycle
By 64, 9 operations have been performed
Therefore, the result of H operation in 2008 is 1
2. When y > 0, there are two groups: (0,5); (1,4)
When y