y=(2x+3)/(x+1)=[(2x+2)+1]/(x+1)=2+1/(x+1)
所以,函数应该就是证明y=1/（x+1）是减函数.
设x1>x2>-1
则y1－y2=1/(x1+1)－1/(x2+1)=[(x2+1)－(x1+1)]/[(x1+1)(x2+1)]=(x2－x1)/[(x1+1)(x2+1)]
∵(x2－x1)0
∴(x2－x1)/[(x1+1)(x2+1)]