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f(1-x)=1/[2^(1-x)+√2]
上下乘2^x,且2^(1-x)*2^x=2^(1-x+x)=2
所以f(1-x)=2^x/[2+√2*2^x)
所以f(x)+f(1-x)
=1/(2^x +√2)+2^x/(2+√2*2^x)
=√2/(2+√2*2^x)+2^x/(2+√2*2^x)
=(√2+2^x)/(2+√2*2^x)
=(√2/2)*(2+√2*2^x)/(2+√2*2^x)
=√2/2

f(-5)=f(1-6)
f(-4)=f(1-5)
f(-3)=f(1-4)
f(-2)=f(1-3)
f(-1)=f(1-2)
f(0)=f(1-1)
所以f（-5）+f（-4）+……+f（5）+f（6）
=6*√2/2
=3√2