已知函数，，f（x）=log2 1-mx⼀x-1 的图像关于原点对称

解：（1）函数f(x) = log2 [(1 – mx)/(x – 1)]的图像关于原点对称，说明函数f(x)是奇函数，定义域关于原点对称，求f(x)的定义域：(1 – mx)/(x – 1) > 0 => (1 – mx)(x – 1) > 0，所以当且仅当m = -1时，对应f(x) = log2 [(1 + x)/(x – 1)]，定义域是(-∞，-1)∪(1，+∞)，f(x)的图像关于原点对称，符合题意；
（2）f(x)在(1，+∞)上单调递减。证明：f(x) = log2 [(1 + x)/(x – 1)]，x∈(1，+∞)，任取x1 ，x2 ∈(1，+∞)，而且x1 < x2 ，计算f(x2) – f(x1) = log2 [(1 + x2)/(x2 – 1)] – log2 [(1 + x1)/(x1 – 1)] = log2 {[(1 + x2)/(x2 – 1)] / [(1 + x1)/(x1 – 1)]} = log2 {[(x2 + 1)(x1 – 1)] / [(x2 – 1)(x1 + 1)]} = log2 [(x1x2 – x2 + x1 – 1) / (x1x2 + x2 – x1 – 1)]①，因为x1 < x2 ，x2 – x1 > 0 > x1 – x2 ，所以(x1x2 + x2 – x1 – 1) > (x1x2 – x2 + x1 – 1) > 0，所以(x1x2 – x2 + x1 – 1) / (x1x2 + x2 – x1 – 1) ∈(0，1)，进而log2 [(x1x2 – x2 + x1 – 1) / (x1x2 + x2 – x1 – 1)] ∈(-∞，0)，所以f(x2) – f(x1) = log2 [(x1x2 – x2 + x1 – 1) / (x1x2 + x2 – x1 – 1)] < 0，即f(x2) < f(x1)，可知f(x)在(1，+∞)上单调递减。

m=+-1