F(X)=(2^x)/(2^x+√2)=1-√2/(2^x+√2)=y,
设F(X)上任意点A(x,y),关于点P(1/2,1/2)对称的点为B,则B(1-x,1-y)
将x=1-x代入函数,得F(1-X)=1-√2/(2^(1-x)+√2)=1-√2/(2*2^(-x)+√2)
=1-1/(√2*2^(-x)+1)=1-2^x/(√2+2^x)=√2/(√2+2^x)=1-[1-√2/(2^x+√2)]=1-y
即对称点B(1-x,1-y)也在函数F(X)上,∴函数F(X)关于点P(1/2,1/2)对称
由F(1-X)=1-y=1-F(X)得,F(1-X)+F(X)=1
∴ f(1/10)+f(2/10)+……f(8/10)+f(9/10)
=[f(1/10)+f(9/10)]+[f(2/10)+f(8/10)]+...+1/2[f(5/10)+f(5/10)]
=[f(1-9/10)+f(9/10)]+[f(1-8/10)+f(8/10)]+...+1/2[f(1-5/10)+f(5/10)]
=1+1+1+1+1/2
=4.5