∵f(x)=
2
2x+1+sinx,
∴f(x)+f(x)=
2
2x+1+sinx+
2
2−x+1+sin(-x)=
2
2x+1+
2•2x
1+2x=2,
则f(0)=1,f(-2)+f(-1)+f(0)+f(1)+f(2)=2+2+1=5,
故答案为:5.
∵f(x)=
2
2x+1+sinx,
∴f(x)+f(x)=
2
2x+1+sinx+
2
2−x+1+sin(-x)=
2
2x+1+
2•2x
1+2x=2,
则f(0)=1,f(-2)+f(-1)+f(0)+f(1)+f(2)=2+2+1=5,
故答案为:5.