赞 no-see-no-go 幼苗
共回答了21个问题采纳率：90.5% 举报
f(x)=αsinx-2/(2^x+1)+8
f(-x)=αsin(-x)-2/(2^-x+1)+8
=-αsinx-2*2^x/(2^x+1)+8
=-αsinx-2*(2^x+1-1)/(2^x+1)+8
=-αsinx-2+2/(2^x+1)+8
=-[αsinx-2/(2^x+1)]+6
=-[αsinx-2/(2^x+1)+8]+14
=-f(x)+14
故f(2013)=f[-(-2013)]=-f(-2013)+14=-2+14=12
f(-x)=αsin(-x)-2/(2^-x+1)+8
=-αsinx-2*2^x/(2^x+1)+8
=-αsinx-2*(2^x+1-1)/(2^x+1)+8
=-αsinx-2+2/(2^x+1)+8
=-[αsinx-2/(2^x+1)]+6
=-[αsinx-2/(2^x+1)+8]+14
=-f(x)+14
故f(2013)=f[-(-2013)]=-f(-2013)+14=-2+14=12
1年前 追问
2
分子分母都乘以2^x： 2/(2^-x+1)=2*2^x/[2^x(2^-x+1)] =2*2^x/(1+2^x) =2*2^x/(2^x+1)
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