函数f(x)对于任意x1,x2∈R,恒有f(x1+x2)=f(x1)f(x2),若f(1)=根号下2,则f(6)=?
人气:410 ℃ 时间:2019-09-29 03:27:14
解答
f(6)=f(1+5)=f(1)f(5)
f(5)=f(1+4)=f(1)f(4)
f(4)=f(1+3)=f(1)f(3)
f(3)=f(1+2)=f(1)f(2)
f(2)=f(1+1)=f(1)f(1)
所以 f(6)=f(1)的六次方=8
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