f(x)=aInx+2÷(x+1)(a∈R)a=1f(x)=Inx+2÷(x+1)(a∈R)f'(x)=1/x+(-2)/(x+1)²f'(x)=1/x-2/(x+1)²1/x-2/(x+1)²＞0单调递增x/x²-2/(x+1)²＞0[x(x+1)²-2x²]/x²(x+1)²＞0x+...f'(x)=a/x+(-2)/(x+1)²f'(x)=a/x-2/(x+1)²a/x-2/(x+1)²<0单调递减ax/x²-2/(x+1)²<0[ax(x+1)²-2x²]/x²(x+1)²<0ax(x+1)²-2x²<0两边除以x^2得ax+1/x<2-2a只要2-2a大于ax+1/x的最大值则满足条件，ax+1/x>=2根号a,,所以有2倍根号a<2-2a,解得：a>根号5/2+3/2或a<根号5/2-3/2谢谢。