f(1-x)
=1/[2^(1-x)+√2]
上下乘2^x,且2^(1-x)*2^x=2
所以f(1-x)=2^x/[2+√2*2^x]
上下除以√2
f(1-x)=(√2/2)*2^x/(2^x+√2)
所以f(x)+f(1-x)
=[(√2/2)*2^x+1]/(2^x+√2)
=(√2/2)*(2^x+√2)/(2^x+√2)
=√2/2
所以f(-5)+f(6)=f(1-6)+f(6)=√2/2
f(-4)+f(5)=√2/2
f(-3)+f(4)=√2/2
f(-2)+f(3)=√2/2
f(-1)+f(2)=√2/2
f(0)+f(1)=√2/2
原式=3√2