f(x)=(x^2+2x+1+x+1-1)/(x+1)=((x+1)^2+(x+1)-1)/(x+1)=(x+1)-1/(x+1)+1x>-1 x+1>0 设x+1=k>0(x+1)-1/(x+1)+1=k-1/k+1因为函数y=x和y=-1/x在[0,+无穷)上都单调递增 所以函数y=x-1/x在[0,+无穷)上单调递增且y∈R 所以...