2的4n+1次方除以(4的2n+1次方加16的n次方的和)

更新时间：2018-03-23 17:53:33 122次访问

2^(4n+1)/[4^(2n)+16^n] 

=2*2^(4n)/[2*4^(2n)] 

=2^(4n)/2^(4n) 

=1 

（负27）的负15次方乘以（负9）的二十次方除以（负3）的负7次方 

=(-3^-45)*3^40/(-3^7) 

=3^(-45+40-7) 

=3的负12次方

2^[4n+1]/(4^[2n+1】+16^n)

分母=2^[4n+2】+2^4n

2^(4n+1)/[4^(2n+1)+16^n]

=2^(4n)*2/[4^(2n)*4+16^n]

=16^n*2/(16^n*4+16^n)

=2*16^n/(5*16^n)

=2/5

2^（4n+1)/[4^(2n+1)+16^n]

=2*2^4n/[2^(4n+2)+2^2n]

=2*2^4n/2^4n(2^2+1)

=2/(2^2+1)

=2/5

答案应该是1~

2^(4n+1)/[4^(2n+1)+16^n]

=2^(4n+1）/【2^(2*(2n+1)+2^(4n)]

=2^4n*2/2^4n(2^2+1)

=2/5