f(x)=(2-x)/(x+1)=(3-x-1)/(x+1)=3/(x+1) -1
∵x+1在(-1.+∞)上单调增
∴3/(x+1)在(-1.+∞)上单调减
∴f(x)=在(-1.+∞)上单调减
如果用定义证明：
令-1＜x1＜x2
f(x2)-f(x1)=3/(x2+1) -1-3/(x1+1) +1
=3[1/(x2+1) -1/(x1+1) ]
=3(x1+1-x2-1) / [(x2+1)(x1+1) ]
=3(x1-x2) / [(x2+1)(x1+1) ]
∵(x1-x2)＜0,[(x2+1)(x1+1) ＞0
∴3(x1-x2) / [(x2+1)(x1+1) ] ＜0
即：f(x2)-f(x1)＜0,f(x2)＜f(x1),得证.