两道数学题``??

来源：百度知道 编辑：UC知道 时间：2024/05/26 19:18:26
1,计算:-2^100*0.5^100*(-1)^2003-1/2
2,已知2^m=3,2^n=4,求2^3m+2n的值?

1:-2^100*0.5^100*(-1)^2003-1/2 =(-2*0.5)^100*(-1)^2003 -1/2
=(-1)^2103 -1/2= -3/2

2 2^3m=(2^m)^3=27
2^n=4得到 n=2
2^3m+2n=27+4=31

1:0.5
2：易得n=2
2^3m+2n=(2^m)^3+4=31

1.1/2
2.432

1. -2^100*0.5^100*(-1)^2003-1/2
=-(2*0.5)^100*(-1)^2003-1/2
=-(-1)^(2003)-1/2
=-(-1)-1/2
=1-1/2
=1/2

2. 2^n=4,则n=更号4=2
2^3m+2n
=(2^m)^3+2*2
=3^3+2*2
=27+4
=31

1,计算:-2^100*0.5^100*(-1)^2003-1/2
=(-2*0.5)^100*(-1)-1/2
=1*(-1)-1/2
=-1-1/2
=-2/3
2,已知2^m=3,2^n=4,求2^3m+2n的值?
2^(3m+2n)
=2^3m*3^2n
=(2^m)^3*(2^n)^2
=3^3*4^2
=432