设集合M={-1,0,1},N={2,3,4,5,6},从M到N的映射f满足条件:对每个x属于M,都有x+f(x)+xf(x)为奇数,那么映射个数为
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/11 04:36:16
![设集合M={-1,0,1},N={2,3,4,5,6},从M到N的映射f满足条件:对每个x属于M,都有x+f(x)+xf(x)为奇数,那么映射个数为](/uploads/image/z/10228789-37-9.jpg?t=%E8%AE%BE%E9%9B%86%E5%90%88M%3D%7B-1%2C0%2C1%7D%2CN%3D%7B2%2C3%2C4%2C5%2C6%7D%2C%E4%BB%8EM%E5%88%B0N%E7%9A%84%E6%98%A0%E5%B0%84f%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6%EF%BC%9A%E5%AF%B9%E6%AF%8F%E4%B8%AAx%E5%B1%9E%E4%BA%8EM%2C%E9%83%BD%E6%9C%89x%2Bf%28x%29%2Bxf%28x%29%E4%B8%BA%E5%A5%87%E6%95%B0%2C%E9%82%A3%E4%B9%88%E6%98%A0%E5%B0%84%E4%B8%AA%E6%95%B0%E4%B8%BA)
xQAN@KZZn wKCt3 !A-RY
j jLL5$Ma\ffڮ qf2{GOW6ZY&^mf̈́KT2Dˮ`ixfPCcc m+y ?đN\nHɨ0T$e(b8ANÐEϧf5% ߍSP䑛]fwjNV]VOϋeG
设集合M={-1,0,1},N={2,3,4,5,6},从M到N的映射f满足条件:对每个x属于M,都有x+f(x)+xf(x)为奇数,那么映射个数为
设集合M={-1,0,1},N={2,3,4,5,6},从M到N的映射f满足条件:对每个x属于M,都有x+f(x)+xf(x)为奇数,那么映射个数为
设集合M={-1,0,1},N={2,3,4,5,6},从M到N的映射f满足条件:对每个x属于M,都有x+f(x)+xf(x)为奇数,那么映射个数为
50个
因为f(x)即为其映射,取N中的数值.
x为偶数即x=0时,f(x)=f(0)=3或5时才满足x+f(x)+xf(x)为奇数,即有两种映射可能.
x为奇数即x=-1或1时,f(x)取N中任何数都满足x+f(x)+xf(x)为奇数,即每个数有5种映射可能.
综述,有2*5*5=50个映射.
有疑问可给我发信息.
2