∵f（x）=ax+1x+2=a(x+2)+1−2ax+2＝a+1−2ax+2，∴要使函数f（x）在x∈[-1，+∞）上递减，则1-2a＞0，此时a＜12，要使f（x）不恒为负，即f（x）=ax+1x+2≥0在∈[-1，+∞）有解，当a=0时，f（x）=ax+1x+2=1x+2，此时...