Question

已知t是正整数，且满足（3^t+1),2^t-3^t(2^t+1)=6^3。求t的值

已知2^m=16，3^n=27，4^m乘（-0.25^n）加上（m-n）^2004的值 急求 要过程

Answer 1

第一题显然有问题，不用足够的括号很容易产生误解。估计是[3^(t+1)]*2^t - (3^t)*[2^(t+1)] = 6³
[3^(t+1)]*2^t - (3^t)*[2^(t+1)]
= 3*(3^t)*(2^t) = (3^t)*(2^t)*2
= 3*6^t - 2*6^t
= 6^t = 6³
t = 3

2^m = 16, m = 4
3^n = 27, n = 3
(4^m)(-0.25^n) + (m-n)^2004
= (4^4)[-(1/4)^3] + (4 - 3)^2004
= -(4^4)(1/4^3) + 1^2004
= -4 + 1
= -3