已知函数的定义域儿R,对m,n∈R,恒有f(m+n)=f(m)+f(n)-1且f(-1/2)=0,当x＞-1/2时,f(

任取a、b∈R,且a-1/2
∴f(b)=f(b-a)+f(a)-1
∵f(-1/2)=0
∴f(b)=f(b-a)+f(a)-1+f(-1/2)
∵f(b-a)+f(-1/2)-1=f(b-a-1/2)
∴f(b)=f(b-a-1/2)+f(a)
f(b)-f(a)=f(b-a-1/2)
∵b-a-1/2>-1/2
∴f(b-a-1/2)>0
∴f(b)-f(a)>0
f(b)>f(a)
∴f(x)在R上单调递增