1. [3^(n-1)]^3 *[3^(-3n+1)]^2÷ 9^2n
2. [ 2^(3n+2)-8^(n-1)]/{2*[2^(n-1)]^3}
F3 maths ?
2016-09-23 9:27 am
回答 (2)
2016-09-23 11:44 am
✔ 最佳答案
1. [3^(n－1)]^3 *[3^(－3n+1)]^2/ 9^(2n)=3^(3n－3)*3^(2－6n/3^(4n)
=3^(3n－3+2-6n-4n)
=3^(－7n－1)
2. [2^(3n+2)－8^(n－1)]/{2*[2^(n－1)]^3}
=[2^(3n+2)－2^(3n－3)]/[2*2^(3n－3)]
=2^(3n－3)*(2^5－1)/[2*2^(3n－3)]
=31/2
2016-10-22 11:37 am
1
[3^(n-1)]^3 *[3^(-3n+1)]^2÷ 9^2n
=[3^(3n-3)][3^(-6n+2)]/[3^(4n)]
=[3^(3n-3)]/[3^(4n)]
=1/[3^(7n+1)
2
[ 2^(3n+2)-8^(n-1)]/{2*[2^(n-1)]^3}
={[2^(3n+2)]-[2^(3n-3)]}/{2[2^(3n-3)]
=[2^(3n)][2^(2)-2^(-3)]/[2^(3n-2)]
=[(2^2)-(2^-3)]/(2^-2)
=31/2
[3^(n-1)]^3 *[3^(-3n+1)]^2÷ 9^2n
=[3^(3n-3)][3^(-6n+2)]/[3^(4n)]
=[3^(3n-3)]/[3^(4n)]
=1/[3^(7n+1)
2
[ 2^(3n+2)-8^(n-1)]/{2*[2^(n-1)]^3}
={[2^(3n+2)]-[2^(3n-3)]}/{2[2^(3n-3)]
=[2^(3n)][2^(2)-2^(-3)]/[2^(3n-2)]
=[(2^2)-(2^-3)]/(2^-2)
=31/2
收錄日期: 2021-04-18 15:39:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160923012721AAfmkcB