设k∈Z,2^-2k + 2^-2k-1 - 2^-2k+1 = ( )2^(-2k) +2^(-2k-1) -2^(-2k+1) =2^(-2k)+2^(-2k)*2^(-1)-2^(-2k)*2 =2^(-2k)*(1+1/2-2) =-1/2*2^(-2k) =-2^(-2k-1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 01:37:09
![设k∈Z,2^-2k + 2^-2k-1 - 2^-2k+1 = ( )2^(-2k) +2^(-2k-1) -2^(-2k+1) =2^(-2k)+2^(-2k)*2^(-1)-2^(-2k)*2 =2^(-2k)*(1+1/2-2) =-1/2*2^(-2k) =-2^(-2k-1)](/uploads/image/z/11264504-32-4.jpg?t=%E8%AE%BEk%E2%88%88Z%2C2%5E-2k+%EF%BC%8B+2%5E-2k-1+%EF%BC%8D+2%5E-2k%2B1+%3D+%28+%EF%BC%892%5E%28-2k%29+%2B2%5E%28-2k-1%29+-2%5E%28-2k%2B1%29+%3D2%5E%28-2k%29%2B2%5E%28-2k%29%2A2%5E%28-1%29-2%5E%28-2k%29%2A2+%3D2%5E%28-2k%29%2A%281%2B1%2F2-2%29+%3D-1%2F2%2A2%5E%28-2k%29+%3D-2%5E%28-2k-1%29)
x){n_(8]l{,]C 6TU ;4|MmCPSA2mPYM-mUhij5Z0mu&$<n+!PeH8*p~͠XC<;kҌ
设k∈Z,2^-2k + 2^-2k-1 - 2^-2k+1 = ( )2^(-2k) +2^(-2k-1) -2^(-2k+1) =2^(-2k)+2^(-2k)*2^(-1)-2^(-2k)*2 =2^(-2k)*(1+1/2-2) =-1/2*2^(-2k) =-2^(-2k-1)
设k∈Z,2^-2k + 2^-2k-1 - 2^-2k+1 = ( )
2^(-2k) +2^(-2k-1) -2^(-2k+1)
=2^(-2k)+2^(-2k)*2^(-1)-2^(-2k)*2
=2^(-2k)*(1+1/2-2)
=-1/2*2^(-2k)
=-2^(-2k-1)
设k∈Z,2^-2k + 2^-2k-1 - 2^-2k+1 = ( )2^(-2k) +2^(-2k-1) -2^(-2k+1) =2^(-2k)+2^(-2k)*2^(-1)-2^(-2k)*2 =2^(-2k)*(1+1/2-2) =-1/2*2^(-2k) =-2^(-2k-1)
1
设k∈Z,2^-2k +2^-2k-1 -2^-2k+1等于?
设k∈Z,2^-2k + 2^-2k-1 - 2^-2k+1 = ( )2^(-2k) +2^(-2k-1) -2^(-2k+1) =2^(-2k)+2^(-2k)*2^(-1)-2^(-2k)*2 =2^(-2k)*(1+1/2-2) =-1/2*2^(-2k) =-2^(-2k-1)
设U=Z,A={x|x=2k,k∈Z}.B={x|x=2k+1,k∈Z},求CUA,CUB
设U=Z,A={x|x+2k,k∈Z},B={x|x=2k+1,k∈Z},求CuA,CuB
设 A={x|x=6k+2,k∈Z} B={x|x=3k-1,k∈Z},C={x|2k,k∈Z},判断ABC之间的关系
设A={x|x=2k-1,k∈Z}.B={x|x=2k,k∈z}
设A={x|x=2k,k∈Z},B={x|x=2k-1,k∈Z},C={x|x=2(k+1),k∈Z}D={x|x=2k+1,k∈Z},其中相等的集合是
设全集U=Z A={xlx=2k+1,k∈Z},B={xlx=2k+1,k∈Z}.求CuB打错了应该是;设全集U=Z A={xlx=2k+1,k∈Z},B={xlx=2k+1,k∈Z}.求CuA
设U=Z,A={x|x=2k,k∈Z};B={x|x=2k+1,k∈Z}求CuA,CuB?设U=Z,A={x|x=2k,k∈Z};B={x|x=2k+1,k∈Z}求CuA,CuB?有急用.
{x|x=3k+1,k∈z}={x|x=3k-2,k∈z}
设U=Z.A={X/X=2K,K属于Z},B={X/X=2K+1,K属于Z},求CuA,CuB
设x0是方程4-x^2=log2底 x的一个解 若x0∈(k,k+1) k∈Z求K
设函数f(x)=lgx+2x-8的零点x0属于(k,k+1),k∈Z,则k=?
设x是实数,且满足等式(x/2)+1/2x=cosθ,则实数θ等于A.2kπ(k∈Z)B.(2k+1)π(k∈Z)C.kπ(k∈Z)D.(1/2)kπ(k∈Z)
设A=X|X=2K+1,K∈Z,B=X|X=2K,K∈Z,则A∩B=( ),A∪B=(
设全集U={x|x=2k,k∈Z},集合A={x=4k,k∈Z},求CuA
设A={x|x=2k,k∈z}B={x|x=2k+1,k∈z}C={x|x=4k+1,k∈z}设a∈A,b∈B判断 a+b=?
设集合A={x|x=3k+1,k属于Z},B={x|x=2k,k属于Z},则A交B=?