推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)可不可以存在规律:f(x+a)=1÷f(x) => f(x+2a)=f(x)推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)啊,
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/22 00:15:16
![推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)可不可以存在规律:f(x+a)=1÷f(x) => f(x+2a)=f(x)推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)啊,](/uploads/image/z/15084848-56-8.jpg?t=%E6%8E%A8%E8%AF%81%EF%BC%9Af%28x%2B1%29%3D1%C3%B7f%28x%29+%3D%3Ef+%28x%2B2%29%3Df%28x%29+f%28x%2B2%29%3D1%C3%B7f%28x%29+%3D%3E+f%28x%2B4%29%3Df%28x%29%E5%8F%AF%E4%B8%8D%E5%8F%AF%E4%BB%A5%E5%AD%98%E5%9C%A8%E8%A7%84%E5%BE%8B%EF%BC%9Af%28x%2Ba%29%3D1%C3%B7f%28x%29+%3D%3E+f%28x%2B2a%29%3Df%28x%29%E6%8E%A8%E8%AF%81%EF%BC%9Af%28x%2B1%29%3D1%C3%B7f%28x%29+%3D%3Ef+%28x%2B2%29%3Df%28x%29+f%28x%2B2%29%3D1%C3%B7f%28x%29+%3D%3E+f%28x%2B4%29%3Df%28x%29%E5%95%8A%EF%BC%8C)
推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)可不可以存在规律:f(x+a)=1÷f(x) => f(x+2a)=f(x)推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)啊,
推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)
可不可以存在规律:f(x+a)=1÷f(x) => f(x+2a)=f(x)
推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)啊,
推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)可不可以存在规律:f(x+a)=1÷f(x) => f(x+2a)=f(x)推证:f(x+1)=1÷f(x) =>f (x+2)=f(x) f(x+2)=1÷f(x) => f(x+4)=f(x)啊,
由f(x+1)=1/f(x)——①
得f(x+2)=f[(x+1)+1]=1/f(x+1)——②
将①代人②得f(x+2)=1/(1/f(x))=f(x)
同理可推证:f(x+2)=1÷f(x) => f(x+4)=f(x)
f(x+a)=1÷f(x) => f(x+2a)=f(x)
f(x+a)=1/f(x)——③
得f[(x+a)+a]=1/f(x+a)——④
③代人④f(x+2a)=f(x)
注:[ ]的写法是我发明的,只为看的顺眼
如果在f(x)没有零点的情况下 是可以推出的
可以!