求 f(x) + f(1-x)
f(x) = 1/(2^x + √2）
f(1-x)
= 1/[2^(1-x) + √2） .(分子、分母同时乘以 2^x )
= 2^x/(2 + √2 * 2^x) .（分母中提取出 √2）
= (2^x/√2) * (1/√2 + 2^x)
= (2^x/√2) * f(x)
f(x) + f(1-x)
= (1+ 2^x/√2) * f(x)
=[ (√2 + 2^x)/√2 ] * f(x)
= [1/√2*f(x)] * f(x)
= 1/√2
f(-5)+f(-4)+…+f(0)+…+f(5)+f(6)
= [f(-5) + f(6)] + [f(-4) + f(5)] + [f(-3) + f(4)] + [f(-2) + f(3)] + [f(-1) + f(2)] + [f(0) + f(1)]
= 6/√2
=3√2