验证就免了,随便代几个>1的数进去看一下,是不是x越取得大,y的值越小,就可以了.下面给出证明过程：
证：
设1f(x2)-f(x1)
=(2x2+3)/(x2-1) -(2x1+3)/(x1-1)
=[(2x2+3)(x1-1)-(2x1+3)(x2-1)]/[(x2-1)(x1-1)]
=(2x1x2-2x2+3x1-3-2x1x2+2x1-3x2+3)/[(x2-1)(x1-1)]
=(-5x2+5x1)/[(x2-1)(x1-1)]
=-5(x2-x1)/[(x2-1)(x1-1)]
x1>1x2>1x2-1>0x1-1>0
x2>x1x2-x1>0
-5<0
-5(x2-x1)/[(x2-1)(x1-1)]<0
f(x2)-f(x1)<0
f(x2)函数是减函数.