math_jj_347

jaze

考试的时候遇见这个， 估计会不淡定

20/2^5 =1/2^m + 1/2^n
5*2^(-3) = 2^(-m) + 2^(-n) à 5*2^(m+n-3)= 2^m + 2^n à 2^(m+n)  /  (2^m+ 2^n) = 8/5

= 2^3/(2^2+ 2^0) = 2^(3 + k) / (2^(2+k) + 2^k)

so  3 +k = m + n; 2 + k = m   k = n  --> 3 + k = 2 + 2k  --> k= 1 m = 3 n = 1

16/10 = 2^4/(2^3 + 2^1) = 8/5  --> mn = 3

can someone give a easier method to solve this problem ?  thanks

sdcar2010

20/2^5 =1/2^m + 1/2^n

So 5/8 = 1/2^m + 1/2^n

Since 5/8 = 1/8 +1/2, then mn=1*3 = 3

jaze

考试的时候要是能这么从容就好了啊

sdcar2010

考试的时候要是能这么从容就好了啊

-- by 会员 jaze (2011/4/10 22:32:30)

Isn't that what JJ is for?